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;
}

};