Combination Sum

题目描述：

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.
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]
]

解题思路：使用回溯法

AC代码如下：

class Solution{
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target){
vector<vector<int>> ans;
vector<int> selectNum;
sort(candidates.begin(), candidates.end());
help(candidates, target, 0, selectNum, ans);
set<vector<int>> uniqueAns(ans.begin(), ans.end());
ans = vector<vector<int>>(uniqueAns.begin(), uniqueAns.end());
return ans;
}

void help(const vector<int>& candidates, int target, int cur_sum, vector<int> selectNum, vector<vector<int>>& ans){
for (int i = 0; i < candidates.size(); ++i){
int sum = cur_sum + candidates[i];
if (sum > target) return;
else if (sum == target){
selectNum.push_back(candidates[i]);
sort(selectNum.begin(), selectNum.end());
ans.push_back(selectNum);
return;
}
else{
selectNum.push_back(candidates[i]);
help(candidates, target, sum, selectNum, ans);
selectNum.pop_back();
}
}
}
};