LeetCode 0039 - Combination Sum

Reck Zhang

发布于 2021-08-11 10:36:26

1890

发布于 2021-08-11 10:36:26

文章被收录于专栏：Reck Zhang

Combination Sum

Desicription

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

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

For example, given candidate set [2, 3, 6, 7] and target 7,

A solution set is:

[
  [7],
  [2, 2, 3]
]

Solution

class Solution {
private:
    vector<int> tmp;
    vector<vector<int> > res;
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, 0);
        return res;
    }

    void dfs(vector<int>& candidates, int target, int index){
        if(index == candidates.size())
            return ;
        int sum = 0;
        for(int i = 0; i < tmp.size(); i++)
            sum += tmp[i];
        if(sum == target){
            res.push_back(tmp);
            return ;
        }
        else if(sum > target)
            return ;
        else{
            for(int i = index; i < candidates.size(); i++){
                tmp.push_back(candidates[i]);
                dfs(candidates, target, i);
                tmp.pop_back();
            }
        }
    }
};

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原始发表：2017-11-11，如有侵权请联系 cloudcommunity@tencent.com 删除

combinations