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[feat] add ruby code - chapter dynamic programming (#1378)
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=begin
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File: climbing_stairs_backtrack.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 回溯 ###
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def backtrack(choices, state, n, res)
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# 当爬到第 n 阶时,方案数量加 1
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res[0] += 1 if state == n
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# 遍历所有选择
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for choice in choices
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# 剪枝:不允许越过第 n 阶
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next if state + choice > n
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# 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res)
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end
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# 回退
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end
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### 爬楼梯:回溯 ###
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def climbing_stairs_backtrack(n)
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choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
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state = 0 # 从第 0 阶开始爬
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res = [0] # 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res)
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res.first
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 9
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res = climbing_stairs_backtrack(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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end
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=begin
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File: climbing_stairs_constraint_dp.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 带约束爬楼梯:动态规划 ###
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def climbing_stairs_backtrack(n)
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return 1 if n == 1 || n == 2
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# 初始化 dp 表,用于存储子问题的解
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dp = Array.new(n + 1) { Array.new(3, 0) }
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# 初始状态:预设最小子问题的解
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dp[1][1], dp[1][2] = 1, 0
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dp[2][1], dp[2][2] = 0, 1
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# 状态转移:从较小子问题逐步求解较大子问题
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for i in 3...(n + 1)
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dp[i][1] = dp[i - 1][2]
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dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
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end
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dp[n][1] + dp[n][2]
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 9
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res = climbing_stairs_backtrack(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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end
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=begin
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File: climbing_stairs_dfs.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 搜索 ###
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def dfs(i)
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# 已知 dp[1] 和 dp[2] ,返回之
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return i if i == 1 || i == 2
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# dp[i] = dp[i-1] + dp[i-2]
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dfs(i - 1) + dfs(i - 2)
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end
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### 爬楼梯:搜索 ###
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def climbing_stairs_dfs(n)
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dfs(n)
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 9
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res = climbing_stairs_dfs(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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end
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=begin
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File: climbing_stairs_dfs_mem.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 记忆化搜索 ###
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def dfs(i, mem)
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# 已知 dp[1] 和 dp[2] ,返回之
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return i if i == 1 || i == 2
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# 若存在记录 dp[i] ,则直接返回之
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return mem[i] if mem[i] != -1
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# dp[i] = dp[i-1] + dp[i-2]
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count = dfs(i - 1, mem) + dfs(i - 2, mem)
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# 记录 dp[i]
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mem[i] = count
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end
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### 爬楼梯:记忆化搜索 ###
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def climbing_stairs_dfs_mem(n)
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# mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
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mem = Array.new(n + 1, -1)
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dfs(n, mem)
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 9
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res = climbing_stairs_dfs_mem(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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end
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40
codes/ruby/chapter_dynamic_programming/climbing_stairs_dp.rb
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40
codes/ruby/chapter_dynamic_programming/climbing_stairs_dp.rb
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=begin
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File: climbing_stairs_dp.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 爬楼梯:动态规划 ###
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def climbing_stairs_dp(n)
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return n if n == 1 || n == 2
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# 初始化 dp 表,用于存储子问题的解
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dp = Array.new(n + 1, 0)
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# 初始状态:预设最小子问题的解
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dp[1], dp[2] = 1, 2
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# 状态转移:从较小子问题逐步求解较大子问题
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(3...(n + 1)).each { |i| dp[i] = dp[i - 1] + dp[i - 2] }
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dp[n]
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end
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### 爬楼梯:空间优化后的动态规划 ###
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def climbing_stairs_dp_comp(n)
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return n if n == 1 || n == 2
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a, b = 1, 2
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(3...(n + 1)).each { a, b = b, a + b }
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b
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 9
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res = climbing_stairs_dp(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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res = climbing_stairs_dp_comp(n)
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puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
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end
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65
codes/ruby/chapter_dynamic_programming/coin_change.rb
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65
codes/ruby/chapter_dynamic_programming/coin_change.rb
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=begin
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File: coin_change.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 零钱兑换:动态规划 ###
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def coin_change_dp(coins, amt)
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n = coins.length
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_MAX = amt + 1
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# 初始化 dp 表
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dp = Array.new(n + 1) { Array.new(amt + 1, 0) }
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# 状态转移:首行首列
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(1...(amt + 1)).each { |a| dp[0][a] = _MAX }
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# 状态转移:其余行和列
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for i in 1...(n + 1)
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for a in 1...(amt + 1)
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if coins[i - 1] > a
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# 若超过目标金额,则不选硬币 i
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dp[i][a] = dp[i - 1][a]
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else
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# 不选和选硬币 i 这两种方案的较小值
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dp[i][a] = [dp[i - 1][a], dp[i][a - coins[i - 1]] + 1].min
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end
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end
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end
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dp[n][amt] != _MAX ? dp[n][amt] : -1
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end
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### 零钱兑换:空间优化后的动态规划 ###
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def coin_change_dp_comp(coins, amt)
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n = coins.length
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_MAX = amt + 1
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# 初始化 dp 表
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dp = Array.new(amt + 1, _MAX)
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dp[0] = 0
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# 状态转移
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for i in 1...(n + 1)
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# 正序遍历
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for a in 1...(amt + 1)
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if coins[i - 1] > a
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# 若超过目标金额,则不选硬币 i
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dp[a] = dp[a]
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else
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# 不选和选硬币 i 这两种方案的较小值
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dp[a] = [dp[a], dp[a - coins[i - 1]] + 1].min
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end
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end
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end
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dp[amt] != _MAX ? dp[amt] : -1
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end
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### Driver Code ###
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if __FILE__ == $0
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coins = [1, 2, 5]
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amt = 4
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# 动态规划
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res = coin_change_dp(coins, amt)
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puts "凑到目标金额所需的最少硬币数量为 #{res}"
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# 空间优化后的动态规划
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res = coin_change_dp_comp(coins, amt)
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puts "凑到目标金额所需的最少硬币数量为 #{res}"
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end
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63
codes/ruby/chapter_dynamic_programming/coin_change_ii.rb
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63
codes/ruby/chapter_dynamic_programming/coin_change_ii.rb
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=begin
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File: coin_change_ii.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 零钱兑换 II:动态规划 ###
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def coin_change_ii_dp(coins, amt)
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n = coins.length
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# 初始化 dp 表
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dp = Array.new(n + 1) { Array.new(amt + 1, 0) }
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# 初始化首列
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(0...(n + 1)).each { |i| dp[i][0] = 1 }
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# 状态转移
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for i in 1...(n + 1)
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for a in 1...(amt + 1)
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if coins[i - 1] > a
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# 若超过目标金额,则不选硬币 i
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dp[i][a] = dp[i - 1][a]
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else
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# 不选和选硬币 i 这两种方案之和
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dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]]
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end
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end
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end
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dp[n][amt]
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end
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### 零钱兑换 II:空间优化后的动态规划 ###
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def coin_change_ii_dp_comp(coins, amt)
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n = coins.length
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# 初始化 dp 表
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dp = Array.new(amt + 1, 0)
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dp[0] = 1
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# 状态转移
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for i in 1...(n + 1)
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# 正序遍历
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for a in 1...(amt + 1)
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if coins[i - 1] > a
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# 若超过目标金额,则不选硬币 i
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dp[a] = dp[a]
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else
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# 不选和选硬币 i 这两种方案之和
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dp[a] = dp[a] + dp[a - coins[i - 1]]
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end
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end
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end
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dp[amt]
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end
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### Driver Code ###
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if __FILE__ == $0
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coins = [1, 2, 5]
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amt = 5
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# 动态规划
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res = coin_change_ii_dp(coins, amt)
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puts "凑出目标金额的硬币组合数量为 #{res}"
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# 空间优化后的动态规划
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res = coin_change_ii_dp_comp(coins, amt)
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puts "凑出目标金额的硬币组合数量为 #{res}"
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end
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115
codes/ruby/chapter_dynamic_programming/edit_distance.rb
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115
codes/ruby/chapter_dynamic_programming/edit_distance.rb
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=begin
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File: edit_distance.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 编辑距离:暴力搜索 ###
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def edit_distance_dfs(s, t, i, j)
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# 若 s 和 t 都为空,则返回 0
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return 0 if i == 0 && j == 0
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# 若 s 为空,则返回 t 长度
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return j if i == 0
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# 若 t 为空,则返回 s 长度
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return i if j == 0
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# 若两字符相等,则直接跳过此两字符
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return edit_distance_dfs(s, t, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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insert = edit_distance_dfs(s, t, i, j - 1)
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delete = edit_distance_dfs(s, t, i - 1, j)
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replace = edit_distance_dfs(s, t, i - 1, j - 1)
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# 返回最少编辑步数
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[insert, delete, replace].min + 1
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end
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def edit_distance_dfs_mem(s, t, mem, i, j)
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# 若 s 和 t 都为空,则返回 0
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return 0 if i == 0 && j == 0
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# 若 s 为空,则返回 t 长度
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return j if i == 0
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# 若 t 为空,则返回 s 长度
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return i if j == 0
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# 若已有记录,则直接返回之
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return mem[i][j] if mem[i][j] != -1
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# 若两字符相等,则直接跳过此两字符
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return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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insert = edit_distance_dfs_mem(s, t, mem, i, j - 1)
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delete = edit_distance_dfs_mem(s, t, mem, i - 1, j)
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replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# 记录并返回最少编辑步数
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mem[i][j] = [insert, delete, replace].min + 1
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end
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### 编辑距离:动态规划 ###
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def edit_distance_dp(s, t)
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n, m = s.length, t.length
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dp = Array.new(n + 1) { Array.new(m + 1, 0) }
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# 状态转移:首行首列
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(1...(n + 1)).each { |i| dp[i][0] = i }
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(1...(m + 1)).each { |j| dp[0][j] = j }
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# 状态转移:其余行和列
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for i in 1...(n + 1)
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for j in 1...(m +1)
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if s[i - 1] == t[j - 1]
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# 若两字符相等,则直接跳过此两字符
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dp[i][j] = dp[i - 1][j - 1]
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else
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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dp[i][j] = [dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]].min + 1
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end
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end
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end
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dp[n][m]
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end
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### 编辑距离:空间优化后的动态规划 ###
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def edit_distance_dp_comp(s, t)
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n, m = s.length, t.length
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dp = Array.new(m + 1, 0)
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# 状态转移:首行
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(1...(m + 1)).each { |j| dp[j] = j }
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# 状态转移:其余行
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for i in 1...(n + 1)
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# 状态转移:首列
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leftup = dp.first # 暂存 dp[i-1, j-1]
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dp[0] += 1
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# 状态转移:其余列
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for j in 1...(m + 1)
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temp = dp[j]
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if s[i - 1] == t[j - 1]
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# 若两字符相等,则直接跳过此两字符
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dp[j] = leftup
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else
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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dp[j] = [dp[j - 1], dp[j], leftup].min + 1
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end
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leftup = temp # 更新为下一轮的 dp[i-1, j-1]
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end
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end
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dp[m]
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end
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### Driver Code ###
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if __FILE__ == $0
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s = 'bag'
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t = 'pack'
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n, m = s.length, t.length
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# 暴力搜索
|
||||
res = edit_distance_dfs(s, t, n, m)
|
||||
puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
|
||||
|
||||
# 记忆化搜索
|
||||
mem = Array.new(n + 1) { Array.new(m + 1, -1) }
|
||||
res = edit_distance_dfs_mem(s, t, mem, n, m)
|
||||
puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
|
||||
|
||||
# 动态规划
|
||||
res = edit_distance_dp(s, t)
|
||||
puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
|
||||
|
||||
# 空间优化后的动态规划
|
||||
res = edit_distance_dp_comp(s, t)
|
||||
puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
|
||||
end
|
99
codes/ruby/chapter_dynamic_programming/knapsack.rb
Normal file
99
codes/ruby/chapter_dynamic_programming/knapsack.rb
Normal file
|
@ -0,0 +1,99 @@
|
|||
=begin
|
||||
File: knapsack.rb
|
||||
Created Time: 2024-05-29
|
||||
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
|
||||
=end
|
||||
|
||||
### 0-1 背包:暴力搜索 ###
|
||||
def knapsack_dfs(wgt, val, i, c)
|
||||
# 若已选完所有物品或背包无剩余容量,则返回价值 0
|
||||
return 0 if i == 0 || c == 0
|
||||
# 若超过背包容量,则只能选择不放入背包
|
||||
return knapsack_dfs(wgt, val, i - 1, c) if wgt[i - 1] > c
|
||||
# 计算不放入和放入物品 i 的最大价值
|
||||
no = knapsack_dfs(wgt, val, i - 1, c)
|
||||
yes = knapsack_dfs(wgt, val, i - 1, c - wgt[i - 1]) + val[i - 1]
|
||||
# 返回两种方案中价值更大的那一个
|
||||
[no, yes].max
|
||||
end
|
||||
|
||||
### 0-1 背包:记忆化搜索 ###
|
||||
def knapsack_dfs_mem(wgt, val, mem, i, c)
|
||||
# 若已选完所有物品或背包无剩余容量,则返回价值 0
|
||||
return 0 if i == 0 || c == 0
|
||||
# 若已有记录,则直接返回
|
||||
return mem[i][c] if mem[i][c] != -1
|
||||
# 若超过背包容量,则只能选择不放入背包
|
||||
return knapsack_dfs_mem(wgt, val, mem, i - 1, c) if wgt[i - 1] > c
|
||||
# 计算不放入和放入物品 i 的最大价值
|
||||
no = knapsack_dfs_mem(wgt, val, mem, i - 1, c)
|
||||
yes = knapsack_dfs_mem(wgt, val, mem, i - 1, c - wgt[i - 1]) + val[i - 1]
|
||||
# 记录并返回两种方案中价值更大的那一个
|
||||
mem[i][c] = [no, yes].max
|
||||
end
|
||||
|
||||
### 0-1 背包:动态规划 ###
|
||||
def knapsack_dp(wgt, val, cap)
|
||||
n = wgt.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(n + 1) { Array.new(cap + 1, 0) }
|
||||
# 状态转移
|
||||
for i in 1...(n + 1)
|
||||
for c in 1...(cap + 1)
|
||||
if wgt[i - 1] > c
|
||||
# 若超过背包容量,则不选物品 i
|
||||
dp[i][c] = dp[i - 1][c]
|
||||
else
|
||||
# 不选和选物品 i 这两种方案的较大值
|
||||
dp[i][c] = [dp[i - 1][c], dp[i - 1][c - wgt[i - 1]] + val[i - 1]].max
|
||||
end
|
||||
end
|
||||
end
|
||||
dp[n][cap]
|
||||
end
|
||||
|
||||
### 0-1 背包:空间优化后的动态规划 ###
|
||||
def knapsack_dp_comp(wgt, val, cap)
|
||||
n = wgt.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(cap + 1, 0)
|
||||
# 状态转移
|
||||
for i in 1...(n + 1)
|
||||
# 倒序遍历
|
||||
for c in cap.downto(1)
|
||||
if wgt[i - 1] > c
|
||||
# 若超过背包容量,则不选物品 i
|
||||
dp[c] = dp[c]
|
||||
else
|
||||
# 不选和选物品 i 这两种方案的较大值
|
||||
dp[c] = [dp[c], dp[c - wgt[i - 1]] + val[i - 1]].max
|
||||
end
|
||||
end
|
||||
end
|
||||
dp[cap]
|
||||
end
|
||||
|
||||
### Driver Code ###
|
||||
if __FILE__ == $0
|
||||
wgt = [10, 20, 30, 40, 50]
|
||||
val = [50, 120, 150, 210, 240]
|
||||
cap = 50
|
||||
n = wgt.length
|
||||
|
||||
# 暴力搜索
|
||||
res = knapsack_dfs(wgt, val, n, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
|
||||
# 记忆化搜索
|
||||
mem = Array.new(n + 1) { Array.new(cap + 1, -1) }
|
||||
res = knapsack_dfs_mem(wgt, val, mem, n, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
|
||||
# 动态规划
|
||||
res = knapsack_dp(wgt, val, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
|
||||
# 空间优化后的动态规划
|
||||
res = knapsack_dp_comp(wgt, val, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
end
|
|
@ -0,0 +1,39 @@
|
|||
=begin
|
||||
File: min_cost_climbing_stairs_dp.rb
|
||||
Created Time: 2024-05-29
|
||||
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
|
||||
=end
|
||||
|
||||
### 爬楼梯最小代价:动态规划 ###
|
||||
def min_cost_climbing_stairs_dp(cost)
|
||||
n = cost.length - 1
|
||||
return cost[n] if n == 1 || n == 2
|
||||
# 初始化 dp 表,用于存储子问题的解
|
||||
dp = Array.new(n + 1, 0)
|
||||
# 初始状态:预设最小子问题的解
|
||||
dp[1], dp[2] = cost[1], cost[2]
|
||||
# 状态转移:从较小子问题逐步求解较大子问题
|
||||
(3...(n + 1)).each { |i| dp[i] = [dp[i - 1], dp[i - 2]].min + cost[i] }
|
||||
dp[n]
|
||||
end
|
||||
|
||||
# 爬楼梯最小代价:空间优化后的动态规划
|
||||
def min_cost_climbing_stairs_dp_comp(cost)
|
||||
n = cost.length - 1
|
||||
return cost[n] if n == 1 || n == 2
|
||||
a, b = cost[1], cost[2]
|
||||
(3...(n + 1)).each { |i| a, b = b, [a, b].min + cost[i] }
|
||||
b
|
||||
end
|
||||
|
||||
### Driver Code ###
|
||||
if __FILE__ == $0
|
||||
cost = [0, 1, 10, 1, 1, 1, 10, 1, 1, 10, 1]
|
||||
puts "输入楼梯的代价列表为 #{cost}"
|
||||
|
||||
res = min_cost_climbing_stairs_dp(cost)
|
||||
puts "爬完楼梯的最低代价为 #{res}"
|
||||
|
||||
res = min_cost_climbing_stairs_dp_comp(cost)
|
||||
puts "爬完楼梯的最低代价为 #{res}"
|
||||
end
|
93
codes/ruby/chapter_dynamic_programming/min_path_sum.rb
Normal file
93
codes/ruby/chapter_dynamic_programming/min_path_sum.rb
Normal file
|
@ -0,0 +1,93 @@
|
|||
=begin
|
||||
File: min_path_sum.rb
|
||||
Created Time: 2024-05-29
|
||||
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
|
||||
=end
|
||||
|
||||
### 最小路径和:暴力搜索 ###
|
||||
def min_path_sum_dfs(grid, i, j)
|
||||
# 若为左上角单元格,则终止搜索
|
||||
return grid[i][j] if i == 0 && j == 0
|
||||
# 若行列索引越界,则返回 +∞ 代价
|
||||
return Float::INFINITY if i < 0 || j < 0
|
||||
# 计算从左上角到 (i-1, j) 和 (i, j-1) 的最小路径代价
|
||||
up = min_path_sum_dfs(grid, i - 1, j)
|
||||
left = min_path_sum_dfs(grid, i, j - 1)
|
||||
# 返回从左上角到 (i, j) 的最小路径代价
|
||||
[left, up].min + grid[i][j]
|
||||
end
|
||||
|
||||
### 最小路径和:记忆化搜索 ###
|
||||
def min_path_sum_dfs_mem(grid, mem, i, j)
|
||||
# 若为左上角单元格,则终止搜索
|
||||
return grid[0][0] if i == 0 && j == 0
|
||||
# 若行列索引越界,则返回 +∞ 代价
|
||||
return Float::INFINITY if i < 0 || j < 0
|
||||
# 若已有记录,则直接返回
|
||||
return mem[i][j] if mem[i][j] != -1
|
||||
# 左边和上边单元格的最小路径代价
|
||||
up = min_path_sum_dfs_mem(grid, mem, i - 1, j)
|
||||
left = min_path_sum_dfs_mem(grid, mem, i, j - 1)
|
||||
# 记录并返回左上角到 (i, j) 的最小路径代价
|
||||
mem[i][j] = [left, up].min + grid[i][j]
|
||||
end
|
||||
|
||||
### 最小路径和:动态规划 ###
|
||||
def min_path_sum_dp(grid)
|
||||
n, m = grid.length, grid.first.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(n) { Array.new(m, 0) }
|
||||
dp[0][0] = grid[0][0]
|
||||
# 状态转移:首行
|
||||
(1...m).each { |j| dp[0][j] = dp[0][j - 1] + grid[0][j] }
|
||||
# 状态转移:首列
|
||||
(1...n).each { |i| dp[i][0] = dp[i - 1][0] + grid[i][0] }
|
||||
# 状态转移:其余行和列
|
||||
for i in 1...n
|
||||
for j in 1...m
|
||||
dp[i][j] = [dp[i][j - 1], dp[i - 1][j]].min + grid[i][j]
|
||||
end
|
||||
end
|
||||
dp[n -1][m -1]
|
||||
end
|
||||
|
||||
### 最小路径和:空间优化后的动态规划 ###
|
||||
def min_path_sum_dp_comp(grid)
|
||||
n, m = grid.length, grid.first.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(m, 0)
|
||||
# 状态转移:首行
|
||||
dp[0] = grid[0][0]
|
||||
(1...m).each { |j| dp[j] = dp[j - 1] + grid[0][j] }
|
||||
# 状态转移:其余行
|
||||
for i in 1...n
|
||||
# 状态转移:首列
|
||||
dp[0] = dp[0] + grid[i][0]
|
||||
# 状态转移:其余列
|
||||
(1...m).each { |j| dp[j] = [dp[j - 1], dp[j]].min + grid[i][j] }
|
||||
end
|
||||
dp[m - 1]
|
||||
end
|
||||
|
||||
### Driver Code ###
|
||||
if __FILE__ == $0
|
||||
grid = [[1, 3, 1, 5], [2, 2, 4, 2], [5, 3, 2, 1], [4, 3, 5, 2]]
|
||||
n, m = grid.length, grid.first.length
|
||||
|
||||
# 暴力搜索
|
||||
res = min_path_sum_dfs(grid, n - 1, m - 1)
|
||||
puts "从左上角到右下角的做小路径和为 #{res}"
|
||||
|
||||
# 记忆化搜索
|
||||
mem = Array.new(n) { Array.new(m, - 1) }
|
||||
res = min_path_sum_dfs_mem(grid, mem, n - 1, m -1)
|
||||
puts "从左上角到右下角的做小路径和为 #{res}"
|
||||
|
||||
# 动态规划
|
||||
res = min_path_sum_dp(grid)
|
||||
puts "从左上角到右下角的做小路径和为 #{res}"
|
||||
|
||||
# 空间优化后的动态规划
|
||||
res = min_path_sum_dp_comp(grid)
|
||||
puts "从左上角到右下角的做小路径和为 #{res}"
|
||||
end
|
61
codes/ruby/chapter_dynamic_programming/unbounded_knapsack.rb
Normal file
61
codes/ruby/chapter_dynamic_programming/unbounded_knapsack.rb
Normal file
|
@ -0,0 +1,61 @@
|
|||
=begin
|
||||
File: unbounded_knapsack.rb
|
||||
Created Time: 2024-05-29
|
||||
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
|
||||
=end
|
||||
|
||||
### 完全背包:动态规划 ###
|
||||
def unbounded_knapsack_dp(wgt, val, cap)
|
||||
n = wgt.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(n + 1) { Array.new(cap + 1, 0) }
|
||||
# 状态转移
|
||||
for i in 1...(n + 1)
|
||||
for c in 1...(cap + 1)
|
||||
if wgt[i - 1] > c
|
||||
# 若超过背包容量,则不选物品 i
|
||||
dp[i][c] = dp[i - 1][c]
|
||||
else
|
||||
# 不选和选物品 i 这两种方案的较大值
|
||||
dp[i][c] = [dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1]].max
|
||||
end
|
||||
end
|
||||
end
|
||||
dp[n][cap]
|
||||
end
|
||||
|
||||
### 完全背包:空间优化后的动态规划 ##3
|
||||
def unbounded_knapsack_dp_comp(wgt, val, cap)
|
||||
n = wgt.length
|
||||
# 初始化 dp 表
|
||||
dp = Array.new(cap + 1, 0)
|
||||
# 状态转移
|
||||
for i in 1...(n + 1)
|
||||
# 正序遍历
|
||||
for c in 1...(cap + 1)
|
||||
if wgt[i -1] > c
|
||||
# 若超过背包容量,则不选物品 i
|
||||
dp[c] = dp[c]
|
||||
else
|
||||
# 不选和选物品 i 这两种方案的较大值
|
||||
dp[c] = [dp[c], dp[c - wgt[i - 1]] + val[i - 1]].max
|
||||
end
|
||||
end
|
||||
end
|
||||
dp[cap]
|
||||
end
|
||||
|
||||
### Driver Code ###
|
||||
if __FILE__ == $0
|
||||
wgt = [1, 2, 3]
|
||||
val = [5, 11, 15]
|
||||
cap = 4
|
||||
|
||||
# 动态规划
|
||||
res = unbounded_knapsack_dp(wgt, val, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
|
||||
# 空间优化后的动态规划
|
||||
res = unbounded_knapsack_dp_comp(wgt, val, cap)
|
||||
puts "不超过背包容量的最大物品价值为 #{res}"
|
||||
end
|
Loading…
Reference in a new issue