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 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, a1a2 ≤ … ≤ 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]

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;
    }
};


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