leetcode-Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums toT.

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.

For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]

[2, 2, 3]

注意：元素可以重复选择。

class Solution {    void helper(vector<int> &base,vector<vector<int> > &res,vector<int>& can,int left){        int max=base.empty()?INT_MIN:base.back(); //选择暂存数组base的最大值        for(auto &e:can){            if(e<max)      //因为暂存数组必须是升序，所以如果e小于末尾元素则不考虑                continue;            if(e==left){   //递归终止的条件，left是剩余的和                base.push_back(e);                res.push_back(base);                  base.pop_back();            }else if(e<left){                base.push_back(e);                helper(base,res,can,left-e);                base.pop_back();            }else           //因为candidates是递增数组                break;        }    }public:    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        vector<vector<int> > res;        if(candidates.empty()||target<=0)            return res;        vector<int> base;        sort(candidates.begin(),candidates.end());        helper(base,res,candidates,target);        return res;    }};