LeetCode--Combination Sum

题目：

Given a set of candidate numbers (C) (without duplicates) 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.
• 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]]

解读：给予一组不重复的正整数数组C，和一个正整数T。从C数组中找出所有数字组合使得和为T，所用的元素可以重复。最终结果中的数组不能重复。

思维：采用递归的方法。设一个整数sum存储当前预备结果的数组的元素和，如果sum>target则返回并把预备结果数组最后一个元素弹出；如果sum==target则把预备结果数组放入最终的结果中，返回并弹出最后元素。用low来存储搜索的数组开始的下标。

代码如图：

class Solution {public:       void combinesum(int target, int& sum, vector<int>& candidates,vector<int>& solution,                                   int low, vector<vector<int>>& result){        if(sum > target) return;        if(sum == target) result.push_back(solution);                for(int i = low; i < candidates.size(); i++){            sum += candidates[i];            solution.push_back(candidates[i]);            combinesum(target, sum, candidates, solution, i , result);            sum -= candidates[i];            solution.pop_back();        }    }    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        int sum = 0;        int low = 0;        vector<vector<int>> result;        vector<int> solution;               combinesum(target,sum, candidates,solution, low, result);        return result;    }     };