leetcode习题解答：39. Combination Sum

难度：Medium

描述：

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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;    }};