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:
target) will be positive integers.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]
]
思路：回溯法
1 class Solution { 2 void combinationSum1(vector<int> candidates, int target, vector<vector<int> > &res, vector<int> &v, int len, int begin) { 3 if (target == 0) { 4 res.push_back(v); 5 return ; 6 } 7 for (int i = begin; (i < len) && target >= candidates[i]; i++) { 8 v.push_back(candidates[i]); 9 combinationSum1(candidates, target - candidates[i], res, v, len, i); 10 v.pop_back(); 11 } 12 13 } 14 public: 15 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 16 sort(candidates.begin(), candidates.end()); 17 vector<vector<int> > res; 18 vector<int> v; 19 int len = candidates.size(); 20 combinationSum1(candidates, target, res, v, len, 0); 21 return res; 22 } 23 };
原文：https://www.cnblogs.com/qinduanyinghua/p/11429513.html