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3f4220de81
* preorder, inorder, postorder -> pre-order, in-order, post-order * Bug fixes * Bug fixes * Update what_is_dsa.md * Sync zh and zh-hant versions * Sync zh and zh-hant versions. * Update performance_evaluation.md and time_complexity.md * Add @khoaxuantu to the landing page. * Sync zh and zh-hant versions * Add @ khoaxuantu to the landing page of zh-hant and en versions.
51 lines
1.5 KiB
Ruby
51 lines
1.5 KiB
Ruby
=begin
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File: subset_sum_ii.rb
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Created Time: 2024-05-22
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 回溯演算法：子集和 II ###
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def backtrack(state, target, choices, start, res)
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# 子集和等於 target 時，記錄解
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if target.zero?
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res << state.dup
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return
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end
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# 走訪所有選擇
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# 剪枝二：從 start 開始走訪，避免生成重複子集
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# 剪枝三：從 start 開始走訪，避免重複選擇同一元素
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for i in start...choices.length
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# 剪枝一：若子集和超過 target ，則直接結束迴圈
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# 這是因為陣列已排序，後邊元素更大，子集和一定超過 target
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break if target - choices[i] < 0
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# 剪枝四：如果該元素與左邊元素相等，說明該搜尋分支重複，直接跳過
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next if i > start && choices[i] == choices[i - 1]
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# 嘗試：做出選擇，更新 target, start
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state << choices[i]
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# 進行下一輪選擇
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backtrack(state, target - choices[i], choices, i + 1, res)
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# 回退：撤銷選擇，恢復到之前的狀態
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state.pop
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end
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end
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### 求解子集和 II ###
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def subset_sum_ii(nums, target)
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state = [] # 狀態（子集）
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nums.sort! # 對 nums 進行排序
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start = 0 # 走訪起始點
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res = [] # 結果串列（子集串列）
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backtrack(state, target, nums, start, res)
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res
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end
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### Driver Code ###
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if __FILE__ == $0
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nums = [4, 4, 5]
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target = 9
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res = subset_sum_ii(nums, target)
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puts "輸入陣列 nums = #{nums}, target = #{target}"
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puts "所有和等於 #{target} 的子集 res = #{res}"
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end
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