39. Combination Sum

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]
]

和为目标数的组合，数可以重复使用

C++：

 1 class Solution {
 2 public:
 3     vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
 4         vector<vector<int>> res ;
 5         vector<int> temp ;
 6         backtracking(candidates,res,temp,target,0) ;
 7         return res ;
 8     }
 9     
10     void backtracking(vector<int> candidates ,vector<vector<int>>& res,vector<int>& temp,int target , int start){
11         if(target == 0){
12             res.push_back(temp) ;
13             return ;
14         }
15         for(int i = start ; i < candidates.size() ; i++){
16             if (candidates[i] <= target){
17                 temp.push_back(candidates[i]) ;
18                 backtracking(candidates,res,temp,target-candidates[i],i) ;
19                 temp.pop_back() ;
20             }
21         }
22     }
23 };

 

转载于:https://www.cnblogs.com/mengchunchen/p/10266937.html