Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

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最新推荐文章于 2025-05-25 18:32:47 发布

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本文探讨了给定一组候选数和目标数时如何找出所有不重复的组合，使得组合内的数之和等于目标数。使用深度优先搜索(DFS)算法，并确保每个数仅被使用一次。通过具体实例展示了解决方案。

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

Each number in C 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.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

方法：DFS

class Solution {
private:
    void find(vector<vector<int>>& res, int index,int temp_sum, const int &target,vector<int>& candidates,vector<int>& buffer){
        if(temp_sum == target){
            res.push_back(buffer);
            return;
        }
        else if(temp_sum > target){
            return;
        }


        for(int i = index; i < candidates.size(); ++i){
            if(i != index){
                if(candidates[i] == candidates[i-1]){
                    continue;
                }
            }
            buffer.push_back(candidates[i]);
            find(res,i+1,temp_sum + candidates[i],target,candidates,buffer);
            buffer.pop_back();
        }

    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> buffer;
        sort(candidates.begin(),candidates.end());
        find(res,0,0,target,candidates,buffer);

        return res;

    }
};