DFS（全排列）--求子数组相加和为sum的方案（包含重复元素）

39. Combination Sum

Medium

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

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

Note:

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

Example 1:

Input: candidates = ​​[2,3,6,7], ​​​target = ​​7​​, A solution set is: [ [7], [2,2,3] ]

Example 2:

Input: candidates = [2,3,5]​​, ​​target = 8, A solution set is: [   [2,2,2,2],   [2,3,3],   [3,5] ]

class Solution {
public:
    void backtracking(vector<vector<int>>& result, vector<int> candidates, vector<int> &tmp_result, int index, int rest_target) {
    if (rest_target == 0) { result.push_back(tmp_result); }
        for (int i = index; i < candidates.size() && candidates[i] <= rest_target; i++) {
            tmp_result.push_back(candidates[i]);
            backtracking(result, candidates, tmp_result, i, rest_target - candidates[i]);
            tmp_result.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> tmp_result;
        sort(candidates.begin(), candidates.end());//从小到大排序
        backtracking(result, candidates, tmp_result, 0, target);
        return result;
    }
};