#include <cstdio>
#include <vector>

using namespace std;

int main()
{
 int n;
 while (scanf("%d", &n) != EOF)
 {
  vector<vector<int>> mat(n, vector<int>(n));
  int num = 1;
  for (int step = 0; step < (n >> 1); ++step)
  {
   for (int j = step; j < n - step; ++j) mat[step][j] = num++;
   for (int i = step + 1; i < n - step; ++i) mat[i][n - 1 - step] = num++;
   for (int j = n - 2 - step; j >= step; --j) mat[n - 1 - step][j] = num++;
   for (int i = n - 2 - step; i > step; --i) mat[i][step] = num++;
  }
  if (n & 1) mat[n >> 1][n >> 1] = num;
  for (int i = 0; i < n; ++i)
  {
   for (int j = 0; j < n - 1; ++j)
    printf("%d ", mat[i][j]);
   printf("%d\n", mat[i][n - 1]);
  }
  printf("\n");
 }

 return 0;
}

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_set>
using namespace std;

int main(int argc, char** argv){
	int n;
	while(cin >> n){
		vector<vector<int>> temp(n, vector<int>(n, 0));
		int begin_x = 0, begin_y = 0, end_x = n, end_y = n;
		int dx[4] = {0, 1, 0, -1};
		int dy[4] = {1, 0, -1, 0};
		int flag = 0;
		int x = 0, y = 0;
		int cur = 1;
		while(cur <= n*n){
			if(x >= begin_x && x < end_x && y >= begin_y && y < end_y){
				temp[x][y] = cur;
				cur++;
				x += dx[flag];
				y += dy[flag]; 
			}
			else{
				if(flag == 0) begin_x++;
				else if(flag == 1) end_y--;
				else if(flag == 2) end_x--;
				else if(flag == 3) begin_y++;
				x -= dx[flag];
				y -= dy[flag]; 
				flag++;
				if(flag == 4) flag = 0;
				x += dx[flag];
				y += dy[flag]; 
			}
		}
		for(int i = 0; i < n; i++){
			for(int j = 0; j < n; j++){
				cout << temp[i][j] << " ";
			}
			cout << endl;
		}
		cout << endl;
	}
	return 0;
}

需要判断重复写入,重复写入将导致出错

1 . j+1<n判断是否到达边界,因为res[0][0]已经赋值了

1. i>=1判断是到达边界

   while True:
    try:
        n = int(input())
        res = [[0 for _ in range(n)] for _ in range(n)]
        tot,i,j=1,0,0
        res[0][0] = tot
        while tot<n**2:
            while j+1<n and not res[i][j+1]:
                tot += 1
                j+=1
                res[i][j]=tot
            while i+1<n and not res[i+1][j]:
                tot += 1
                i += 1
                res[i][j]=tot
            while 1<=j and not res[i][j-1]:
                tot += 1
                j -= 1
                res[i][j]=tot
            while i >=1 and not res[i-1][j]:
                tot += 1
                i -= 1
                res[i][j]=tot
        for i in res:
            print(' '.join(map(str,i)))
        print()
    except:break

一个数组记录是否访问过，一个数组记录走法，同类型的题目不管出成什么样都一样

python solution

import sys


def generateMatrix(n):
    res = [[0 for i in range(n)] for j in range(n)]
    count = 1
    rowIndex1 = 0
    rowIndex2 = n - 1
    colIndex1 = n - 1
    colIndex2 = 0
    rowNum = n
    while rowNum >= 1:
        for i in range(colIndex2, colIndex1 + 1):
            res[rowIndex1][i] = count
            count += 1
        rowIndex1 += 1

        for i in range(rowIndex1, rowIndex2 + 1):
            res[i][colIndex1] = count
            count += 1
        colIndex1 -= 1

        for i in range(colIndex1, colIndex2 - 1, -1):
            res[rowIndex2][i] = count
            count += 1
        rowIndex2 -= 1

        for i in range(rowIndex2, rowIndex1 - 1, -1):
            res[i][colIndex2] = count
            count += 1
        colIndex2 += 1
        rowNum -= 2
    return res


for i in sys.stdin.readlines():
    matrix = generateMatrix(int(i))
    for line in matrix:
        print(" ".join(map(str, line)))
    print()

#include <iostream>
#include <vector>
using namespace std;
int main()
{
	int length;
	while (cin >> length)
	{
		vector<vector<int> > data(length, vector<int>(length, 0));
		vector<int> number(length * 2 - 1, length);
		int cnt = 1;
		for (int i = 1; i < 2 * length - 1; i = i + 2)
		{
			number[i] = length - cnt;
			number[i + 1] = length - cnt;
			cnt++;
		}		
		if (length == 1)
		{
			cout << 1 << endl << endl;
			continue;
		}
		else
		{
			cnt = 1;
			int i = 0, j = -1;
			for (int k = 0; k < 2 * length - 1; k++)
			{
				if (k % 4 == 0)//向右
				{
					for (int x = 0; x < number[k]; x++)
					{
						j++;
						data[i][j] = cnt;
						cnt++;
					}
				}
				else if (k % 4 == 1)//向下
				{
					for (int x = 0; x < number[k]; x++)
					{
						i++;
						data[i][j] = cnt;
						cnt++;
					}
				}
				else if (k % 4 == 2)//向左
				{
					for (int x = 0; x < number[k]; x++)
					{
						j--;
						data[i][j] = cnt;
						cnt++;
					}
				}
				else  //向上
				{
					for (int x = 0; x < number[k]; x++)
					{
						i--;
						data[i][j] = cnt;
						cnt++;
					}
				}
			}
		}
		for (vector<vector<int> >::iterator it = data.begin(); it != data.end(); it++)
		{
			vector<int> res = *it;
			for (vector<int>::iterator k = res.begin(); k != res.end(); k++)
			{
				cout << *k;
				if (k != res.end() - 1)
					cout << " ";
				else
					cout << endl;
			}
		}
		cout << endl;
	}
	return 0;
}

import java.util.*;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int n = sc.nextInt();
            int[][] arr = new int[n][n];
            int value = 1;
            //round为需要转的圈数.n为奇数时n/2+1圈,n为偶数时为n/2圈
            for (int round = 0; round < n / 2 + n % 2; round ++ ) {
                for (int i = round; i < n - round; i ++ ) { // 左→右
                    arr[round][i] = value ++ ;
                }
                for (int i = round + 1; i < n - round; i ++ ) { // 上→下
                    arr[i][n - round - 1] = value ++ ;
                }
                for (int i = n - round - 2; i >= round; i -- ) { // 右→左
                    arr[n - round - 1][i] = value ++ ;
                }
                for (int i = n - round - 2; i >= round + 1; i -- ) { // 下→上
                    arr[i][round] = value ++ ;
                }
            }
            for (int i = 0; i < n; i ++ ) {
                for (int j = 0; j < n; j ++ ) {
                    if(j == n - 1) System.out.println(arr[i][j]);
                    else System.out.print(arr[i][j] + " ");
                }
            }
            System.out.println();
        }
    }
}

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int n;
	while (cin >> n && n)
	{
		vector<vector<int> > m(n, vector<int>(n));
		int dir[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };

		int num = 1, sum = n - 1;
		for (int rou = 0; rou < (n + 1) / 2; ++rou)	// 循环第几圈
		{
			int x = rou, y = rou;
			for (int i = 0; i < 4; ++i)				// →↓←↑四个方向循环
			{
				for (int cou = 0; cou < sum; ++cou)	// 每圈有几个数字（减1防止重复打印）
				{
					m[x][y] = num;					// 打印
					++num;
					x += dir[i][0];					// 计算下一个位置
					y += dir[i][1];
				}
			}
			sum -= 2;								// 经过一圈，个数减2
		}
		if (n % 2) m[n / 2][n / 2] = n*n;			// 奇数情况，打印中心位置

		for (int i = 0; i < n; ++i)					// 输出
		{
			for (int j = 0; j < n; ++j)
				cout << (j ? " " : "") << m[i][j];
			cout << endl;
		}
		cout << endl;								// 每组打印一行空行
	}
	return 0;
}

#include "iostream"
using namespace std; 
int main(void){
	int E,S,W,N;//东南西北四个边界	
	int i,j,x,n;
	while(cin>>n){
        N = W = 0;
	E = S = n-1;
	int a[n][n];
	x = 1;
    while(N!=S && W!=E){
 	 for(i=N,j=W;j<=E;) a[i][j++] = x++;
         N++;
         if(N==S && W==E) break;
         for(i=N,j=E;i<=S;) a[i++][j] = x++;
         E--;
         if(N==S && W==E) break;
         for(i=S,j=E;j>=W;) a[i][j--] = x++;
         S--;
         if(N==S && W==E) break;
         for(i=S,j=W;i>=N;) a[i--][j] = x++;
         W++;
         if(N==S && W==E) break;
    }
    a[N][W] = x;
for(i=0;i<=n-1;i++){
	for(j=0;j<=n-2;j++){
	cout<<a[i][j]<<" ";	//千万不要用ends，搞了半天通不过是因为这个。。。
	}
	cout<<a[i][n-1]<<endl;//最后一个直接换行
}
    cout<<endl;
}
    return 0;
}