leetcode习题解答：39. Combination Sum

难度：Medium

描述：

---

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.

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, 

思路：

他要求数组中加起来等于target的组合，这个数可以重复使用。

可以递归地实现，先排序，不断挑可以选的最小的数字。加进一个数组里，看数组和是不是等于target，等于就加到结果中，小于就把它放到数组，进入下一层递归。

代码：

#include <algorithmn>
class Solution {
public:
    int T;
    int cal(vector<int>& filled){
        int res = 0;
        for (int i = 0; i < filled.size(); i++){
            res += filled[i];
        }
        return res;
    }
    void fill(vector<int>& source, vector<int>& filled, vector<vector<int>>& res,int start){
        for (int i = start; i < source.size(); i++){
            if (cal(filled)+source[i] == T){
                filled.push_back(source[i]);
                res.push_back(filled);
                filled.pop_back();
            } else if (cal(filled)+source[i] < T){
                filled.push_back(source[i]);
                fill(source, filled, res, i);
                filled.pop_back();
            }
        }
        
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        T = target;
        sort(candidates.begin(),candidates.end());
        vector<int> filled;
        vector<vector<int>> res;
        fill(candidates,filled,res,0);
        return res;
    }
};