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51 lines
1.4 KiB
Python
51 lines
1.4 KiB
Python
"""
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File: subset_sum_i_naive.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],
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target: int,
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total: int,
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choices: list[int],
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res: list[list[int]],
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):
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"""回溯算法：子集和 I"""
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# 子集和等于 target 时，记录解
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if total == target:
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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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for i in range(len(choices)):
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# 剪枝：若子集和超过 target ，则跳过该选择
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if total + choices[i] > target:
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continue
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# 尝试：做出选择，更新元素和 total
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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, total + choices[i], choices, 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_i_naive(nums: list[int], target: int) -> list[list[int]]:
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"""求解子集和 I（包含重复子集）"""
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state = [] # 状态（子集）
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total = 0 # 子集和
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res = [] # 结果列表（子集列表）
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backtrack(state, target, total, nums, 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 = [3, 4, 5]
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target = 9
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res = subset_sum_i_naive(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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print(f"请注意，该方法输出的结果包含重复集合")
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