[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;
}

}
}

};