2017小红书校园招聘面试经历

小红书是上海的一家基于b2c的电商，虽然没有阿里巴巴的名气但是也是一家处于快速发展的企业，这个公司对算法比较重视，一上来就要求用递归方法写

一个全排列，幸好还记得一些思路，最后还是被我写出来了，一下是当时写的代码：

#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<sstream>
#include<fstream>
#include<cstring>
#include<stack>
#include<algorithm>
using namespace std;

vector<vector<int>>get_all_permutation(vector<int>&data, int end){
vector<vector<int>>re;
vector<vector<int>>tmp;
if (end == 0){
vector<int>temp;
temp.push_back(data[end]);
re.push_back(temp);
return re;
}
re = get_all_permutation(data, end - 1);
for (int i = 0; i<re.size(); i++){
int count = 0;
vector<int>::iterator iter;
int size = re[i].size();
int index = 0;
while (count != size + 1){
iter = re[i].begin();
for (int k = 0; k < index; k++)iter++;
re[i].insert(iter, data[end]);
tmp.push_back(re[i]);
iter = re[i].begin();
for (int k = 0; k < index; k++)iter++;
re[i].erase(iter);
index++;
count++;
}
}
return tmp;
}

int main() {
int num;
vector<int>data;
vector<vector<int>>re;
int count = 1;
while (count){
num = 0;
if (num<1){
cout << 0 << endl;
return 0;
}
for (int i = 1; i <= num; i++)data.push_back(i);
re = get_all_permutation(data, num - 1);
cout << '[';
for (int i = 0; i<re.size(); i++){
cout << '[';
for (int j = 0; j<re[i].size(); j++){
cout << re[i][j];
if (j != re[i].size() - 1)cout << ',';
}
cout << ']';
if (i != re.size() - 1)cout << ',';
}
cout << ']' << endl;
count--;
}
return 0;
}

//Input: 数字 n
//Output: 以二维数组的形式，输出[1,2,...,n]的全排列。
//例如：当n=3时，输出[[1,2,3], [1,3,2], ..., [3,2,1]]

//标准解法：递归。

这种方法比较乱，主要思想是利用插空法，假设我们在a,b,c三个字符的情况下，我们调用函数递归调用返回b,c和c,b那么我们可以将a插入到b,c和c,b的每一位
置，那么对于b,c来说，就会有abc,bac,bca,对于c,b来说，我们就有acb,cab,cba；

但是在正常情况下,我们的输出序列为：

1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1
后来又写了另外一种方法：

#include<iostream>
#include<string>
#include<vector>
#include<memory>
#include<fstream>
using namespace std;

vector<vector<int>> get_all_permutation(vector<int>&data,int begin){
vector<vector<int>>re;
vector<int>tmp;
if (begin == data.size()-2){//这个时候还剩两个元素
tmp.push_back(data[begin]);
tmp.push_back(data[begin+1]);
re.push_back(tmp);
//交换两个数
tmp[0] = tmp[0] ^ tmp[1];
tmp[1] = tmp[0] ^ tmp[1];
tmp[0] = tmp[0] ^ tmp[1];
re.push_back(tmp);
return re;
}
re = get_all_permutation(data,begin+1);
for (int i = 0; i < re.size(); i++)re[i].insert(re[i].begin(), data[begin]);
int num = re.size();
for (int i = 0; i < num; i++){//将0号下标和从1号下标开始的元素交换
for (int j = 1; j < re[i].size(); j++){
tmp = re[i];
tmp[0] = tmp[0] + tmp[j];
tmp[j] = tmp[0] - tmp[j];
tmp[0] = tmp[0] - tmp[j];
re.push_back(tmp);
}
}
return re;
}

int main(){
int num;
vector<int>data;
vector<vector<int>>re;
while (cin>>num){
if (num == 0)return  0;
if (num == 1){ cout << num << endl; continue; }
for (int i = 1; i <= num; i++)data.push_back(i);
re = get_all_permutation(data,0);
cout << '[';
for (int i = 0; i < re.size();i++){
cout << '[';
for (int j = 0; j < re[i].size(); j++){
cout << re[i][j];
if (j != re[i].size() - 1)cout << ',';
}
cout << ']';
}
cout << ']' << endl;
data.clear();
}
}

再用树的结构来写一遍(这种思想用到dfs和回溯方法)：

#include<vector>
#include<memory>
#include<fstream>
#include<stack>
using namespace std;

void get_all_permutation(vector<int>&data,vector<bool>&flag,vector<int>&stk){//begin代表当前的下标
int size = data.size();
if (stk.size() == data.size()){
cout << '[';
for (int i = 0; i < stk.size(); i++){
cout << stk[i];
if (i != stk.size() - 1)cout << ',';
}
cout << ']';
return;
}
for (int i = 0; i < size;i++){
if (flag[i] != true){//可以继续访问
flag[i] = true;
stk.push_back(data[i]);
get_all_permutation(data,flag,stk);
stk.pop_back();
flag[i] = false;
}
}
}

int main(){
int num;
vector<int>data;
vector<vector<int>>re;
vector<bool>flag;
while (cin>>num){
if (num == 0)return  0;
if (num == 1){ cout << num << endl; continue; }
for (int i = 0; i < num; i++)flag.push_back(false);
for (int i = 1; i <= num; i++)data.push_back(i);
cout << '[';
int count = 0;
vector<int>stack;
get_all_permutation(data,flag,stack);
cout << ']' << endl;
data.clear();
flag.clear();
}
}

可以用STL里面的next_permutation函数实现一遍