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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"""回溯演算法：子集和 I"""
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# 子集和等於 target 時，記錄解
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if total == target:
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res.append(list(state))
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return
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# 走訪所有選擇
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for i in range(len(choices)):
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# 剪枝：若子集和超過 target ，則跳過該選擇
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if total + choices[i] > target:
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continue
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# 嘗試：做出選擇，更新元素和 total
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state.append(choices[i])
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# 進行下一輪選擇
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backtrack(state, target, total + choices[i], choices, 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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def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
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"""求解子集和 I（包含重複子集）"""
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state = [] # 狀態（子集）
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total = 0 # 子集和
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res = [] # 結果串列（子集串列）
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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"輸入陣列 nums = {nums}, target = {target}")
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print(f"所有和等於 {target} 的子集 res = {res}")
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print(f"請注意，該方法輸出的結果包含重複集合")
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