[LeetCode]039-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:
思路，先按照由小到大排序，然后依次选值，假设第i个：candidates[i]，选完后target减去candidates[i]，递归下，继续遍历从i往后的值。直到找到与target一样的值为止。

class Solution {public:    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> results;        vector<int> result;        combine(candidates,target,0,results,result);        return results;    }    void combine(vector<int>& candidates, int target,int begin,vector<vector<int>>& results,vector<int>& result)    {        int n = candidates.size();        int i = 0;        for(i = begin;i<n;i++)        {            if(candidates[i] == target)            {                result.push_back(candidates[i]);                results.push_back(result);                if(!result.empty())                {                    result.pop_back();                }                return;            }            else if(candidates[i] < target)            {                int t = target - candidates[i];                result.push_back(candidates[i]);                combine(candidates,t,i,results,result);                if(!result.empty())                    result.pop_back();            }            else            {                return;            }        }    }};