LeetCode39. Combination Sum
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题目链接:
https://leetcode.com/problems/combination-sum/
题目描述:
给出一组候选数字(C)和目标数字(T),找出C中所有的组合,使组合中数字的和为T。C中每个数字在每个组合中可多次使用。组合中的数字降序排列。
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
思路:
排序,再回溯。和之前Subsets做法差不多,只不过这里的数字可以重复使用而已。
代码:
class Solution {public: vector<vector<int>> res; vector<int> cmb; int len; void combinationSum(vector<int>& candidates,int k,int curSum,int target){ if(curSum>target){ return; } else if(curSum==target){ res.push_back(cmb); } else{ for(int i=k;i<len;i++){ cmb.push_back(candidates[i]); combinationSum(candidates,i,curSum+candidates[i],target); cmb.pop_back(); } } } vector<vector<int>> combinationSum(vector<int>& candidates, int target) { len=candidates.size(); if(len==0){ return res; } sort(candidates.begin(),candidates.end()); combinationSum(candidates,0,0,target); return res; }};
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