Combination Sum I

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]

解法：

由题意为set知，集合应该没有重复元素；用回溯法。

注意，每个元素可以取多次，

void  combinationCore(vector<int> &candi,int target,int begin,vector<int> &tempresult,vector<vector<int> > &results){

if(target==0){//target==0,说明已经找到一个可行解

results.push_back(tempresult);

}

else{

int size=candi.size();

for(int i=begin;i<size&&target>=candi[i];++i){

//target>=candi[i],因为同一个元素可以取多次，则i从begin开始，且下一个也是从i开始，

tempresult.push_back(candi[i]);

combinationCore(candi,target-candi[i],i,tempresult,results);

tempresult.pop_back();

}

}

}

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

vector<vector<int>> results;

int size=candidates.size();

if(size==0||target<=0)

return results;

vector<int> temp;

sort(candidates.begin(),candidates.end());//must

combinationCore(candidates,target,0,temp,results);

return results;

}