[leetcode] 39. Combination Sum 解题报告

题目链接：https://leetcode.com/problems/combination-sum/

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] 

思路：一个DFS, 对于每一个数字可以不拿, 或者拿无限次, 因此在做DFS的时候要么不拿当前的数, 要么枚举拿1,2,3等多个, 直到超过了和的大小.

代码如下：

class Solution {public:    void DFS(vector<int>& candidates, int sum, int k, vector<int> vec)    {        if(sum ==0) ans.push_back(vec);        if(k >= candidates.size() || sum <= 0) return;        DFS(candidates, sum, k+1, vec);        for(int i = 1; sum - i*candidates[k]>=0; i++)        {            vec.push_back(candidates[k]);            DFS(candidates, sum-i*candidates[k], k+1, vec);        }    }        vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        DFS(candidates, target, 0, vector<int>{});        return ans;    }private:    vector<vector<int>> ans;};