Combination Sum I

来源：Leetcode

原帖：http://oj.leetcode.com/problems/combination-sum/

题目：
Given a set of candidate numbers (C) 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. Elements in a combination (a1, a2, .. , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). 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: Sort & Recursion.

代码：

 class Solution {  
 public:  
   vector<vector<int>> combinationSum(vector<int> &candidates, int target) {  
     vector<vector<int> > res;  
     sort(candidates.begin(), candidates.end()); // 如果没有要求结果是非下降序列，则sort是非必要的  
     vector<int> com;  
     combinationSumHelper(candidates, target, 0, com, res);  
     return res;  
   }  
   
   void combinationSumHelper(const vector<int> &num, int target, int start, vector<int> &com, vector<vector<int>> &res) {  
     if (target == 0) {  
       res.push_back(com);  
       return;  
     }  
     for (int i = start; i < num.size() && target >= num[i]; ++i) {  
       com.push_back(num[i]);  
       combinationSumHelper(num, target - num[i], i, com, res);  
       com.pop_back();  
     }  
   }  
 };