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48 lines
1.3 KiB
48 lines
1.3 KiB
6 months ago
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=begin
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File: subset_sum_i.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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### 回溯算法：子集和 I ###
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def backtrack(state, target, choices, start, res)
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# 子集和等于 target 时，记录解
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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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# 遍历所有选择
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# 剪枝二：从 start 开始遍历，避免生成重复子集
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for i in start...choices.length
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# 剪枝一：若子集和超过 target ，则直接结束循环
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# 这是因为数组已排序，后边元素更大，子集和一定超过 target
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break if target - choices[i] < 0
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# 尝试：做出选择，更新 target, start
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state << choices[i]
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# 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res)
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# 回退：撤销选择，恢复到之前的状态
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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 I ###
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def subset_sum_i(nums, target)
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state = [] # 状态（子集）
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nums.sort! # 对 nums 进行排序
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start = 0 # 遍历起始点
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res = [] # 结果列表（子集列表）
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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 = [3, 4, 5]
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target = 9
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res = subset_sum_i(nums, target)
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puts "输入数组 = #{nums}, target = #{target}"
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puts "所有和等于 #{target} 的子集 res = #{res}"
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end
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