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60 lines
1.8 KiB
Python
60 lines
1.8 KiB
Python
"""
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File: coin_change.py
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Created Time: 2023-07-10
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Author: Krahets (krahets@163.com)
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"""
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def coin_change_dp(coins: list[int], amt: int) -> int:
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"""零钱兑换：动态规划"""
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n = len(coins)
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MAX = amt + 1
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# 初始化 dp 表
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dp = [[0] * (amt + 1) for _ in range(n + 1)]
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# 状态转移：首行首列
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for a in range(1, amt + 1):
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dp[0][a] = MAX
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# 状态转移：其余行列
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for i in range(1, n + 1):
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超过背包容量，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] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
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return dp[n][amt] if dp[n][amt] != MAX else -1
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def coin_change_dp_comp(coins: list[int], amt: int) -> int:
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"""零钱兑换：状态压缩后的动态规划"""
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n = len(coins)
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MAX = amt + 1
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# 初始化 dp 表
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dp = [MAX] * (amt + 1)
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dp[0] = 0
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# 状态转移
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for i in range(1, n + 1):
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# 正序遍历
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超过背包容量，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] = min(dp[a], dp[a - coins[i - 1]] + 1)
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return dp[amt] if dp[amt] != MAX else -1
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"""Driver Code"""
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if __name__ == "__main__":
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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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print(f"凑到目标金额所需的最少硬币数量为 {res}")
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# 状态压缩后的动态规划
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res = coin_change_dp_comp(coins, amt)
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print(f"凑到目标金额所需的最少硬币数量为 {res}")
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