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53 lines
1.7 KiB
53 lines
1.7 KiB
"""
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File: subset_sum_ii.py
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Created Time: 2023-06-17
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
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):
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"""回溯算法：子集和 II"""
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# 子集和等于 target 时，记录解
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if target == 0:
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res.append(list(state))
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return
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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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# 剪枝三：从 start 开始遍历，避免重复选择同一元素
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for i in range(start, len(choices)):
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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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if target - choices[i] < 0:
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break
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# 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
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if i > start and choices[i] == choices[i - 1]:
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continue
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# 尝试：做出选择，更新 target, start
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state.append(choices[i])
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# 进行下一轮选择
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backtrack(state, target - choices[i], choices, i + 1, 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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def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
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"""求解子集和 II"""
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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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return res
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"""Driver Code"""
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if __name__ == "__main__":
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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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print(f"输入数组 nums = {nums}, target = {target}")
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print(f"所有和等于 {target} 的子集 res = {res}")
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