LeetCode40. Combination Sum II（C++/Python）

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最新推荐文章于 2025-05-19 01:02:14 发布

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本文介绍了一种算法，用于解决给定一组候选数和目标数，找到所有唯一组合的问题，使得候选数之和等于目标数。该算法采用回溯剪枝策略，避免重复组合，适用于正整数集合，确保解决方案集中没有重复的组合。

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Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

解题思路：回溯剪枝，基本思路与上一题基本一致，但是不同的地方在于，本题数组中每个元素只能使用一次，那么首先需要修改的就是dfs中进行递归的时候，cur的值，但是改了之后，发现会出现这种情况，当由两个1的时候，1,7这个解会有两个，所以我们在循环中修改一下条件，相同的元素直接跳过。

C++

class Solution {
public:
    vector<int>tempres;
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>>res;
        dfs(candidates, target, 0, res);
        return res;
    }
    void dfs(vector<int>& candidates, int target, int cur, vector<vector<int>>& res){
        if(target == 0){
            res.push_back(tempres);
            return;
        }
        if(cur >= candidates.size() || target < candidates[cur])
            return;
        for(int i = cur; i < candidates.size();){
            tempres.push_back(candidates[i]);
            dfs(candidates, target - candidates[i], i+1, res);
            tempres.pop_back();
            int temp = candidates[i];
            while(i < candidates.size() && candidates[i] == temp)
                ++ i;
        }
    }
};

Python

class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        self.res, self.tempres = [], []
        def DFS(candidates, target, cur):
            if not target:
                self.res.append(self.tempres[:])
                return 
            if cur == len(candidates) or target < candidates[cur]:
                return
            i = cur
            while i < len(candidates):
                self.tempres.append(candidates[i])
                DFS(candidates, target-candidates[i], i+1)
                del self.tempres[-1]
                temp = candidates[i]
                i += 1
                while i < len(candidates) and candidates[i] == temp:
                    i += 1
        DFS(candidates, target, 0)
        return self.res