39. Combination Sum

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.

• 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]
]

题意：输出能加起来等于target的全部组合。
思路：先排序，递归求所有数字的集合，并判断加起来是否等于target

class Solution {
public:
    //先构建递归函数
    void CombinationSumCore(const vector<int>& candidates,vector<vector<int>>& res,vector<int>& com,int target,int pos){
        if(target==0){   //全是正数，结果不可能是0
            res.push_back(com);
            return;
        }
        for(int i=pos;i<candidates.size();i++){
            if(candidates[i]>target)break;   //如果这个数太大了就放弃。
            com.push_back(candidates[i]);    
            CombinationSumCore(candidates,res,com,target-candidates[i],i);  //递归（每加入一个数，target就变小一点。）
            com.pop_back(); //跳过当前的数，并计算下一个数
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target){
        sort(candidates.begin(),candidates.end());   //排序
        vector<vector<int>> res;   //存放结果
        vector<int> com;    //存放正在计算的数的集合
        CombinationSumCore(candidates,res,com,target,0);
        return res;
    }
};