问题描述：

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.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

基本思路：

回溯法求解。具体请参见代码。

代码：

class Solution {  //C++
public:
    vector<int> record;
    vector<vector<int> >result;
    set<vector<int> > myset;
    
    void addSolution(){
        vector<int> tmp(record.begin(),record.end());
        sort(tmp.begin(),tmp.end());
        if(myset.find(tmp) == myset.end()){
            result.push_back(tmp);
            myset.insert(tmp);   
        }
    }
    void subCombinationSum(vector<int> &cadidates, int target,int bpos){
        if(target ==0 ){
            addSolution();
        }
        
        int size = cadidates.size();
        for(int i = bpos; i < size&&cadidates[i] <= target; i++){
            record.push_back(cadidates[i]);
            subCombinationSum(cadidates,target-cadidates[i],bpos);
            record.pop_back();
        }
    }
    
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        subCombinationSum(candidates,target,0);
        return result;
    }
};

[leetcode]

原文：http://blog.csdn.net/chenlei0630/article/details/42085961