40.组合总和II

backtracking , https://leetcode.cn/problems/combination-sum-ii/

给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用一次。

注意：解集不能包含重复的组合。

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

示例 2:

输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]

提示:

1 <= candidates.length <= 100
1 <= candidates[i] <= 50
1 <= target <= 30

python

from typing import List

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()  
        res = []  
        
        def dfs(start, remain, path):
            if remain == 0:
                res.append(path)  # 找到一个组合，加入结果
                return
            if remain < 0:
                return  
            
            for i in range(start, len(candidates)):
                num = candidates[i]
                # 跳过重复的元素，避免重复的组合
                if i > start and num == candidates[i-1]:
                    continue
                
                dfs(i + 1, remain - num, path + [num])
        
        dfs(0, target, [])
        
        return res

python

from typing import List


class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()
        res = []

        def backtracking(start, remain, path):
            if remain == 0:
                res.append(list(path))  # 找到一个组合，加入结果
                return
            if remain < 0:
                return

            for i in range(start, len(candidates)):
                num = candidates[i]
                # 跳过重复的元素，避免重复的组合
                if i > start and num == candidates[i - 1]:
                    continue
                path.append(num)
                backtracking(i + 1, remain - num, path)
                path.pop()

        backtracking(0, target, [])

        return res

40.组合总和II ​

40.组合总和II