[LeetCode] Combination Sum 回溯

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] 

 

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  Array Backtracking

 

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   一道回溯题目，可以剪枝提升速度。

 

#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;



class Solution {

public:

    vector<vector<int > > ret;

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {

        vector<int > stk;

        ret.clear();

        vector<int > tmp(candidates.begin(),candidates.end());

        sort(tmp.begin(),tmp.end());

        helpFun(tmp,target,0,stk);

        return ret;

    }



    void helpFun(vector<int> & cand,int tar, int idx,vector<int > & stk)

    {

        if(tar<0)   return ;

        if(tar==0){

            ret.push_back(stk);

            return ;

        }

        if(idx==cand.size())    return;

        stk.push_back(cand[idx]);

        helpFun(cand,tar-cand[idx],idx,stk);

        stk.pop_back();

        helpFun(cand,tar,idx+1,stk);

    }

};



int main()

{

    vector<int > cand = {8,7,4,3};

    Solution sol;

    vector<vector<int > > ret=sol.combinationSum(cand,11);

    for(int i =0;i<ret.size();i++){

        for(int j=0;j<ret[i].size();j++)

            cout<<ret[i][j]<<" ";

        cout<<endl;

    }

    return 0;

}