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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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"""回溯演算法：子集和 II"""
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# 子集和等於 target 時，記錄解
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if target == 0:
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res.append(list(state))
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return
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# 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 range(start, len(choices)):
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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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if target - choices[i] < 0:
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break
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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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if i > start and choices[i] == choices[i - 1]:
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continue
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# 嘗試：做出選擇，更新 target, start
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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 - 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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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 = [] # 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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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"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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