Leecode77 组合

77. 组合 - 力扣(LeetCode)

算法复习:回溯

回溯模板:

void func( flag, mem, cur)

resType in(){

func(initflag,initmem,cur)

}

题目分析

  1. 组合具有唯一性
  2. 组合内数字不重复
  3. 各个组合内的数字各不相同,不存在组合不同组合内部数字相同的情况
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class Solution {
public:
void get_combaition(vector<int>& cur, int index, vector<vector<int>>& res,
int count, bool* arr, int n, int k) {
if (count == k) {
res.push_back(cur);
return;
}
if (index > n)
return;

for (int i = index; i <= n; i++) {
if (arr[i] == 0) {
cur.push_back(i);
arr[i] = 1;
count++;
get_combaition(cur, i + 1, res, count, arr, n, k);
cur.pop_back();
arr[i] = 0;
count--;
}
}
}
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> res;
vector<int> cur;
bool arr[n + 1];
memset(arr, 0, n + 1);
get_combaition(cur, 1, res, 0, arr, n, k);
return res;
}
};
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// ACM模式

#include<bits/stdc++.h>
using namespace std;

int n;
int k;


void get_combaition(vector<int> &cur,int index,vector<vector<int>> &res,int count, bool *arr,int n,int k){
if(count==k) {
res.push_back(cur);
return;
}
if(index>n ) return ;

for(int i = index;i<=n;i++){
if(arr[i] == 0){
cur.push_back(i);
arr[i] = 1;
count++;
get_combaition(cur,i+1,res,count,arr,n,k);
cur.pop_back();
arr[i] = 0;
count--;
}
}
}

vector<vector<int>> get_res(int n,int k){
vector<vector<int>> res;
vector<int> cur;
// vector<int> arr(n+1,0);
bool arr[n+1];
memset(arr,0,n+1);
get_combaition(cur,1,res,0,arr,n,k);
return res;
}




int main(){
int n,k;
cin>>n>>k;
// get_res(n,k);
// std::cout<<count;
// 方法二
vector<vector<int>> res = get_res(n,k);
for(auto it:res){
for(auto it2:it){
cout<<it2<<" ";
}
cout<<endl;
}



return 0;
}