# #The way to brush questions with leetcode 40 combination sum II

Given an array candidates and a target number target, find out all combinations of candidates that can make numbers and targets.
Each number in candidates can only be used once in each combination.
Explain:
All numbers, including the target number, are positive integers.
A solution set cannot contain duplicate combinations.

Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
The solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
The solution set is:
[
[1,2,2],
[5]
]

That is to say 39 questions Add one to avoid repetition:

```#include <iostream>
#include <vector>
#include <algorithm>
using  namespace std;
static int i=0;

void recurrent(vector<int>& candidates,int target,int addr,vector<vector<int>>& ans,vector<int>& temp)//addr Is the current traversal position,
{
if(target==0)
{
//        if(temp.size()==1)
//        cout<<temp[0]<<endl;
//        if(temp.size()==3)
//            cout<<temp[0]<<temp[1]<<temp[2]<<endl;
if(find(ans.begin(),ans.end(),temp)==ans.end()) ans.push_back(temp);//Avoid duplication
return;
}
if(target<0) return;
{
target-=candidates[i];
temp.push_back(candidates[i]);
recurrent(candidates,target,i+1,ans,temp);
temp.pop_back();//Restore to continue traversal
target+=candidates[i];
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> ans;
int count=candidates.size();
if(count==0||target<0) return ans;
vector<int> temp;
sort(candidates.begin(),candidates.end());//Sort, avoid repetition
recurrent(candidates,target,0,ans,temp);
return ans;
}

int main() {
vector<int> candidates={2,5,2,1,2};
vector<vector<int>> ans=combinationSum2(candidates,5);
std::cout << ans.size()<< std::endl;
return 0;
}```

Tags: C++

Posted on Wed, 04 Dec 2019 07:01:09 -0500 by capbiker