35 - Combination Sum

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: 递归方法来解决，先排序以确保结果集都是非递减的；解决的关键在于每次递归都从当前位开始遍历，把数添加到sum中去，而不是从当前位的后一位开始，这样能保证重复数字的解决方案存在。

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector< vector<int> >result;
        vector<int>res;
        
        if( candidates.size() <=0 || target <= 0 )
            return result;
            
        sort(candidates.begin(), candidates.end());
        
        int sum = 0;
        generateCom(candidates, result, res, 0, sum, target);
        
        
        return result;
    }
    
    void generateCom(vector<int> &candidates, vector< vector<int> >&result, vector<int>&res, int cur, int &sum,int target)
    {
        if(sum > target)return;
        else if(sum == target)
        {
            result.push_back(res);
            return;
        }
        for(int i = cur; i < candidates.size(); i++)
        {
            sum += candidates[i];
            res.push_back(candidates[i]);
            generateCom(candidates, result, res, i, sum, target);
            res.pop_back();
            sum -= candidates[i];
        }
    }
};