797 lines
20 KiB
Markdown
797 lines
20 KiB
Markdown
<p align="center">
|
||
<a href="https://www.programmercarl.com/xunlian/xunlianying.html" target="_blank">
|
||
<img src="../pics/训练营.png" width="1000"/>
|
||
</a>
|
||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||
|
||
|
||
|
||
# 59.螺旋矩阵II
|
||
|
||
[力扣题目链接](https://leetcode.cn/problems/spiral-matrix-ii/)
|
||
|
||
给定一个正整数 n，生成一个包含 1 到 n^2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。
|
||
|
||
示例:
|
||
|
||
输入: 3
|
||
输出:
|
||
[
|
||
[ 1, 2, 3 ],
|
||
[ 8, 9, 4 ],
|
||
[ 7, 6, 5 ]
|
||
]
|
||
|
||
|
||
## 算法公开课
|
||
|
||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[拿下螺旋矩阵！LeetCode：59.螺旋矩阵II](https://www.bilibili.com/video/BV1SL4y1N7mV)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
|
||
|
||
## 思路
|
||
|
||
这道题目可以说在面试中出现频率较高的题目，**本题并不涉及到什么算法，就是模拟过程，但却十分考察对代码的掌控能力。**
|
||
|
||
要如何画出这个螺旋排列的正方形矩阵呢？
|
||
|
||
相信很多同学刚开始做这种题目的时候，上来就是一波判断猛如虎。
|
||
|
||
结果运行的时候各种问题，然后开始各种修修补补，最后发现改了这里那里有问题，改了那里这里又跑不起来了。
|
||
|
||
大家还记得我们在这篇文章[数组：每次遇到二分法，都是一看就会，一写就废](https://programmercarl.com/0704.二分查找.html)中讲解了二分法，提到如果要写出正确的二分法一定要坚持**循环不变量原则**。
|
||
|
||
而求解本题依然是要坚持循环不变量原则。
|
||
|
||
模拟顺时针画矩阵的过程:
|
||
|
||
* 填充上行从左到右
|
||
* 填充右列从上到下
|
||
* 填充下行从右到左
|
||
* 填充左列从下到上
|
||
|
||
由外向内一圈一圈这么画下去。
|
||
|
||
可以发现这里的边界条件非常多，在一个循环中，如此多的边界条件，如果不按照固定规则来遍历，那就是**一进循环深似海，从此offer是路人**。
|
||
|
||
这里一圈下来，我们要画每四条边，这四条边怎么画，每画一条边都要坚持一致的左闭右开，或者左开右闭的原则，这样这一圈才能按照统一的规则画下来。
|
||
|
||
那么我按照左闭右开的原则，来画一圈，大家看一下：
|
||
|
||
|
||
|
||
这里每一种颜色，代表一条边，我们遍历的长度，可以看出每一个拐角处的处理规则，拐角处让给新的一条边来继续画。
|
||
|
||
这也是坚持了每条边左闭右开的原则。
|
||
|
||
一些同学做这道题目之所以一直写不好，代码越写越乱。
|
||
|
||
就是因为在画每一条边的时候，一会左开右闭，一会左闭右闭，一会又来左闭右开，岂能不乱。
|
||
|
||
代码如下，已经详细注释了每一步的目的，可以看出while循环里判断的情况是很多的，代码里处理的原则也是统一的左闭右开。
|
||
|
||
整体C++代码如下：
|
||
|
||
```CPP
|
||
class Solution {
|
||
public:
|
||
vector<vector<int>> generateMatrix(int n) {
|
||
vector<vector<int>> res(n, vector<int>(n, 0)); // 使用vector定义一个二维数组
|
||
int startx = 0, starty = 0; // 定义每循环一个圈的起始位置
|
||
int loop = n / 2; // 每个圈循环几次，例如n为奇数3，那么loop = 1 只是循环一圈，矩阵中间的值需要单独处理
|
||
int mid = n / 2; // 矩阵中间的位置，例如：n为3， 中间的位置就是(1，1)，n为5，中间位置为(2, 2)
|
||
int count = 1; // 用来给矩阵中每一个空格赋值
|
||
int offset = 1; // 需要控制每一条边遍历的长度，每次循环右边界收缩一位
|
||
int i,j;
|
||
while (loop --) {
|
||
i = startx;
|
||
j = starty;
|
||
|
||
// 下面开始的四个for就是模拟转了一圈
|
||
// 模拟填充上行从左到右(左闭右开)
|
||
for (j; j < n - offset; j++) {
|
||
res[i][j] = count++;
|
||
}
|
||
// 模拟填充右列从上到下(左闭右开)
|
||
for (i; i < n - offset; i++) {
|
||
res[i][j] = count++;
|
||
}
|
||
// 模拟填充下行从右到左(左闭右开)
|
||
for (; j > starty; j--) {
|
||
res[i][j] = count++;
|
||
}
|
||
// 模拟填充左列从下到上(左闭右开)
|
||
for (; i > startx; i--) {
|
||
res[i][j] = count++;
|
||
}
|
||
|
||
// 第二圈开始的时候，起始位置要各自加1， 例如：第一圈起始位置是(0, 0)，第二圈起始位置是(1, 1)
|
||
startx++;
|
||
starty++;
|
||
|
||
// offset 控制每一圈里每一条边遍历的长度
|
||
offset += 1;
|
||
}
|
||
|
||
// 如果n为奇数的话，需要单独给矩阵最中间的位置赋值
|
||
if (n % 2) {
|
||
res[mid][mid] = count;
|
||
}
|
||
return res;
|
||
}
|
||
};
|
||
```
|
||
|
||
* 时间复杂度 O(n^2): 模拟遍历二维矩阵的时间
|
||
* 空间复杂度 O(1)
|
||
|
||
## 类似题目
|
||
|
||
* [54.螺旋矩阵](https://leetcode.cn/problems/spiral-matrix/)
|
||
* [剑指Offer 29.顺时针打印矩阵](https://leetcode.cn/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/)
|
||
|
||
|
||
|
||
|
||
## 其他语言版本
|
||
|
||
### Java：
|
||
|
||
```Java
|
||
class Solution {
|
||
public int[][] generateMatrix(int n) {
|
||
int[][] nums = new int[n][n];
|
||
int startX = 0, startY = 0; // 每一圈的起始点
|
||
int offset = 1;
|
||
int count = 1; // 矩阵中需要填写的数字
|
||
int loop = 1; // 记录当前的圈数
|
||
int i, j; // j 代表列, i 代表行;
|
||
|
||
while (loop <= n / 2) {
|
||
|
||
// 顶部
|
||
// 左闭右开，所以判断循环结束时， j 不能等于 n - offset
|
||
for (j = startY; j < n - offset; j++) {
|
||
nums[startX][j] = count++;
|
||
}
|
||
|
||
// 右列
|
||
// 左闭右开，所以判断循环结束时， i 不能等于 n - offset
|
||
for (i = startX; i < n - offset; i++) {
|
||
nums[i][j] = count++;
|
||
}
|
||
|
||
// 底部
|
||
// 左闭右开，所以判断循环结束时， j != startY
|
||
for (; j > startY; j--) {
|
||
nums[i][j] = count++;
|
||
}
|
||
|
||
// 左列
|
||
// 左闭右开，所以判断循环结束时， i != startX
|
||
for (; i > startX; i--) {
|
||
nums[i][j] = count++;
|
||
}
|
||
startX++;
|
||
startY++;
|
||
offset++;
|
||
loop++;
|
||
}
|
||
if (n % 2 == 1) { // n 为奇数时，单独处理矩阵中心的值
|
||
nums[startX][startY] = count;
|
||
}
|
||
return nums;
|
||
}
|
||
}
|
||
|
||
|
||
```
|
||
|
||
### python3:
|
||
|
||
```python
|
||
class Solution:
|
||
def generateMatrix(self, n: int) -> List[List[int]]:
|
||
nums = [[0] * n for _ in range(n)]
|
||
startx, starty = 0, 0 # 起始点
|
||
loop, mid = n // 2, n // 2 # 迭代次数、n为奇数时，矩阵的中心点
|
||
count = 1 # 计数
|
||
|
||
for offset in range(1, loop + 1) : # 每循环一层偏移量加1，偏移量从1开始
|
||
for i in range(starty, n - offset) : # 从左至右，左闭右开
|
||
nums[startx][i] = count
|
||
count += 1
|
||
for i in range(startx, n - offset) : # 从上至下
|
||
nums[i][n - offset] = count
|
||
count += 1
|
||
for i in range(n - offset, starty, -1) : # 从右至左
|
||
nums[n - offset][i] = count
|
||
count += 1
|
||
for i in range(n - offset, startx, -1) : # 从下至上
|
||
nums[i][starty] = count
|
||
count += 1
|
||
startx += 1 # 更新起始点
|
||
starty += 1
|
||
|
||
if n % 2 != 0 : # n为奇数时，填充中心点
|
||
nums[mid][mid] = count
|
||
return nums
|
||
```
|
||
|
||
版本二：定义四个边界
|
||
```python
|
||
class Solution(object):
|
||
def generateMatrix(self, n):
|
||
if n <= 0:
|
||
return []
|
||
|
||
# 初始化 n x n 矩阵
|
||
matrix = [[0]*n for _ in range(n)]
|
||
|
||
# 初始化边界和起始值
|
||
top, bottom, left, right = 0, n-1, 0, n-1
|
||
num = 1
|
||
|
||
while top <= bottom and left <= right:
|
||
# 从左到右填充上边界
|
||
for i in range(left, right + 1):
|
||
matrix[top][i] = num
|
||
num += 1
|
||
top += 1
|
||
|
||
# 从上到下填充右边界
|
||
for i in range(top, bottom + 1):
|
||
matrix[i][right] = num
|
||
num += 1
|
||
right -= 1
|
||
|
||
# 从右到左填充下边界
|
||
|
||
for i in range(right, left - 1, -1):
|
||
matrix[bottom][i] = num
|
||
num += 1
|
||
bottom -= 1
|
||
|
||
# 从下到上填充左边界
|
||
|
||
for i in range(bottom, top - 1, -1):
|
||
matrix[i][left] = num
|
||
num += 1
|
||
left += 1
|
||
|
||
return matrix
|
||
```
|
||
|
||
### JavaScript:
|
||
|
||
```javascript
|
||
|
||
var generateMatrix = function(n) {
|
||
let startX = startY = 0; // 起始位置
|
||
let loop = Math.floor(n/2); // 旋转圈数
|
||
let mid = Math.floor(n/2); // 中间位置
|
||
let offset = 1; // 控制每一层填充元素个数
|
||
let count = 1; // 更新填充数字
|
||
let res = new Array(n).fill(0).map(() => new Array(n).fill(0));
|
||
|
||
while (loop--) {
|
||
let row = startX, col = startY;
|
||
// 上行从左到右（左闭右开）
|
||
for (; col < n - offset; col++) {
|
||
res[row][col] = count++;
|
||
}
|
||
// 右列从上到下（左闭右开）
|
||
for (; row < n - offset; row++) {
|
||
res[row][col] = count++;
|
||
}
|
||
// 下行从右到左（左闭右开）
|
||
for (; col > startY; col--) {
|
||
res[row][col] = count++;
|
||
}
|
||
// 左列做下到上（左闭右开）
|
||
for (; row > startX; row--) {
|
||
res[row][col] = count++;
|
||
}
|
||
|
||
// 更新起始位置
|
||
startX++;
|
||
startY++;
|
||
|
||
// 更新offset
|
||
offset += 1;
|
||
}
|
||
// 如果n为奇数的话，需要单独给矩阵最中间的位置赋值
|
||
if (n % 2 === 1) {
|
||
res[mid][mid] = count;
|
||
}
|
||
return res;
|
||
};
|
||
|
||
|
||
|
||
```
|
||
|
||
### TypeScript:
|
||
|
||
```typescript
|
||
function generateMatrix(n: number): number[][] {
|
||
let loopNum: number = Math.floor(n / 2);
|
||
const resArr: number[][] = new Array(n).fill(1).map(i => new Array(n));
|
||
let chunkNum: number = n - 1;
|
||
let startX: number = 0;
|
||
let startY: number = 0;
|
||
let value: number = 1;
|
||
let x: number, y: number;
|
||
while (loopNum--) {
|
||
x = startX;
|
||
y = startY;
|
||
while (x < startX + chunkNum) {
|
||
resArr[y][x] = value;
|
||
x++;
|
||
value++;
|
||
}
|
||
while (y < startY + chunkNum) {
|
||
resArr[y][x] = value;
|
||
y++;
|
||
value++;
|
||
}
|
||
while (x > startX) {
|
||
resArr[y][x] = value;
|
||
x--;
|
||
value++;
|
||
}
|
||
while (y > startY) {
|
||
resArr[y][x] = value;
|
||
y--;
|
||
value++;
|
||
}
|
||
startX++;
|
||
startY++;
|
||
chunkNum -= 2;
|
||
}
|
||
if (n % 2 === 1) {
|
||
resArr[startX][startY] = value;
|
||
}
|
||
return resArr;
|
||
};
|
||
```
|
||
|
||
### Go:
|
||
|
||
```go
|
||
package main
|
||
|
||
import "fmt"
|
||
|
||
func main() {
|
||
n := 3
|
||
fmt.Println(generateMatrix(n))
|
||
}
|
||
|
||
func generateMatrix(n int) [][]int {
|
||
startx, starty := 0, 0
|
||
var loop int = n / 2
|
||
var center int = n / 2
|
||
count := 1
|
||
offset := 1
|
||
res := make([][]int, n)
|
||
for i := 0; i < n; i++ {
|
||
res[i] = make([]int, n)
|
||
}
|
||
for loop > 0 {
|
||
i, j := startx, starty
|
||
|
||
//行数不变 列数在变
|
||
for j = starty; j < n-offset; j++ {
|
||
res[startx][j] = count
|
||
count++
|
||
}
|
||
//列数不变是j 行数变
|
||
for i = startx; i < n-offset; i++ {
|
||
res[i][j] = count
|
||
count++
|
||
}
|
||
//行数不变 i 列数变 j--
|
||
for ; j > starty; j-- {
|
||
res[i][j] = count
|
||
count++
|
||
}
|
||
//列不变 行变
|
||
for ; i > startx; i-- {
|
||
res[i][j] = count
|
||
count++
|
||
}
|
||
startx++
|
||
starty++
|
||
offset++
|
||
loop--
|
||
}
|
||
if n%2 == 1 {
|
||
res[center][center] = n * n
|
||
}
|
||
return res
|
||
}
|
||
```
|
||
|
||
```go
|
||
func generateMatrix(n int) [][]int {
|
||
top, bottom := 0, n-1
|
||
left, right := 0, n-1
|
||
num := 1
|
||
tar := n * n
|
||
matrix := make([][]int, n)
|
||
for i := 0; i < n; i++ {
|
||
matrix[i] = make([]int, n)
|
||
}
|
||
for num <= tar {
|
||
for i := left; i <= right; i++ {
|
||
matrix[top][i] = num
|
||
num++
|
||
}
|
||
top++
|
||
for i := top; i <= bottom; i++ {
|
||
matrix[i][right] = num
|
||
num++
|
||
}
|
||
right--
|
||
for i := right; i >= left; i-- {
|
||
matrix[bottom][i] = num
|
||
num++
|
||
}
|
||
bottom--
|
||
for i := bottom; i >= top; i-- {
|
||
matrix[i][left] = num
|
||
num++
|
||
}
|
||
left++
|
||
}
|
||
return matrix
|
||
}
|
||
```
|
||
|
||
### Swift:
|
||
|
||
```swift
|
||
func generateMatrix(_ n: Int) -> [[Int]] {
|
||
var result = [[Int]](repeating: [Int](repeating: 0, count: n), count: n)
|
||
|
||
var startRow = 0
|
||
var startColumn = 0
|
||
var loopCount = n / 2
|
||
let mid = n / 2
|
||
var count = 1
|
||
var offset = 1
|
||
var row: Int
|
||
var column: Int
|
||
|
||
while loopCount > 0 {
|
||
row = startRow
|
||
column = startColumn
|
||
|
||
for c in column ..< startColumn + n - offset {
|
||
result[startRow][c] = count
|
||
count += 1
|
||
column += 1
|
||
}
|
||
|
||
for r in row ..< startRow + n - offset {
|
||
result[r][column] = count
|
||
count += 1
|
||
row += 1
|
||
}
|
||
|
||
for _ in startColumn ..< column {
|
||
result[row][column] = count
|
||
count += 1
|
||
column -= 1
|
||
}
|
||
|
||
for _ in startRow ..< row {
|
||
result[row][column] = count
|
||
count += 1
|
||
row -= 1
|
||
}
|
||
|
||
startRow += 1
|
||
startColumn += 1
|
||
offset += 2
|
||
loopCount -= 1
|
||
}
|
||
|
||
if (n % 2) != 0 {
|
||
result[mid][mid] = count
|
||
}
|
||
return result
|
||
}
|
||
```
|
||
|
||
### Rust:
|
||
|
||
```rust
|
||
impl Solution {
|
||
pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {
|
||
let mut res = vec![vec![0; n as usize]; n as usize];
|
||
let (mut startX, mut startY, mut offset): (usize, usize, usize) = (0, 0, 1);
|
||
let mut loopIdx = n/2;
|
||
let mid: usize = loopIdx as usize;
|
||
let mut count = 1;
|
||
let (mut i, mut j): (usize, usize) = (0, 0);
|
||
while loopIdx > 0 {
|
||
i = startX;
|
||
j = startY;
|
||
|
||
while j < (startY + (n as usize) - offset) {
|
||
res[i][j] = count;
|
||
count += 1;
|
||
j += 1;
|
||
}
|
||
|
||
while i < (startX + (n as usize) - offset) {
|
||
res[i][j] = count;
|
||
count += 1;
|
||
i += 1;
|
||
}
|
||
|
||
while j > startY {
|
||
res[i][j] = count;
|
||
count += 1;
|
||
j -= 1;
|
||
}
|
||
|
||
while i > startX {
|
||
res[i][j] = count;
|
||
count += 1;
|
||
i -= 1;
|
||
}
|
||
|
||
startX += 1;
|
||
startY += 1;
|
||
offset += 2;
|
||
loopIdx -= 1;
|
||
}
|
||
|
||
if(n % 2 == 1) {
|
||
res[mid][mid] = count;
|
||
}
|
||
res
|
||
}
|
||
}
|
||
```
|
||
|
||
### PHP:
|
||
|
||
```php
|
||
class Solution {
|
||
/**
|
||
* @param Integer $n
|
||
* @return Integer[][]
|
||
*/
|
||
function generateMatrix($n) {
|
||
// 初始化数组
|
||
$res = array_fill(0, $n, array_fill(0, $n, 0));
|
||
$mid = $loop = floor($n / 2);
|
||
$startX = $startY = 0;
|
||
$offset = 1;
|
||
$count = 1;
|
||
while ($loop > 0) {
|
||
$i = $startX;
|
||
$j = $startY;
|
||
for (; $j < $startY + $n - $offset; $j++) {
|
||
$res[$i][$j] = $count++;
|
||
}
|
||
for (; $i < $startX + $n - $offset; $i++) {
|
||
$res[$i][$j] = $count++;
|
||
}
|
||
for (; $j > $startY; $j--) {
|
||
$res[$i][$j] = $count++;
|
||
}
|
||
for (; $i > $startX; $i--) {
|
||
$res[$i][$j] = $count++;
|
||
}
|
||
$startX += 1;
|
||
$startY += 1;
|
||
$offset += 2;
|
||
$loop--;
|
||
}
|
||
if ($n % 2 == 1) {
|
||
$res[$mid][$mid] = $count;
|
||
}
|
||
return $res;
|
||
}
|
||
}
|
||
```
|
||
|
||
### C:
|
||
|
||
```c
|
||
int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
|
||
//初始化返回的结果数组的大小
|
||
*returnSize = n;
|
||
*returnColumnSizes = (int*)malloc(sizeof(int) * n);
|
||
//初始化返回结果数组ans
|
||
int** ans = (int**)malloc(sizeof(int*) * n);
|
||
int i;
|
||
for(i = 0; i < n; i++) {
|
||
ans[i] = (int*)malloc(sizeof(int) * n);
|
||
(*returnColumnSizes)[i] = n;
|
||
}
|
||
|
||
//设置每次循环的起始位置
|
||
int startX = 0;
|
||
int startY = 0;
|
||
//设置二维数组的中间值，若n为奇数。需要最后在中间填入数字
|
||
int mid = n / 2;
|
||
//循环圈数
|
||
int loop = n / 2;
|
||
//偏移数
|
||
int offset = 1;
|
||
//当前要添加的元素
|
||
int count = 1;
|
||
|
||
while(loop) {
|
||
int i = startX;
|
||
int j = startY;
|
||
//模拟上侧从左到右
|
||
for(; j < startY + n - offset; j++) {
|
||
ans[startX][j] = count++;
|
||
}
|
||
//模拟右侧从上到下
|
||
for(; i < startX + n - offset; i++) {
|
||
ans[i][j] = count++;
|
||
}
|
||
//模拟下侧从右到左
|
||
for(; j > startY; j--) {
|
||
ans[i][j] = count++;
|
||
}
|
||
//模拟左侧从下到上
|
||
for(; i > startX; i--) {
|
||
ans[i][j] = count++;
|
||
}
|
||
//偏移值每次加2
|
||
offset+=2;
|
||
//遍历起始位置每次+1
|
||
startX++;
|
||
startY++;
|
||
loop--;
|
||
}
|
||
//若n为奇数需要单独给矩阵中间赋值
|
||
if(n%2)
|
||
ans[mid][mid] = count;
|
||
|
||
return ans;
|
||
}
|
||
```
|
||
### Scala:
|
||
|
||
```scala
|
||
object Solution {
|
||
def generateMatrix(n: Int): Array[Array[Int]] = {
|
||
var res = Array.ofDim[Int](n, n) // 定义一个n*n的二维矩阵
|
||
var num = 1 // 标志当前到了哪个数字
|
||
var i = 0 // 横坐标
|
||
var j = 0 // 竖坐标
|
||
|
||
while (num <= n * n) {
|
||
// 向右：当j不越界，并且下一个要填的数字是空白时
|
||
while (j < n && res(i)(j) == 0) {
|
||
res(i)(j) = num // 当前坐标等于num
|
||
num += 1 // num++
|
||
j += 1 // 竖坐标+1
|
||
}
|
||
i += 1 // 下移一行
|
||
j -= 1 // 左移一列
|
||
|
||
// 剩下的都同上
|
||
|
||
// 向下
|
||
while (i < n && res(i)(j) == 0) {
|
||
res(i)(j) = num
|
||
num += 1
|
||
i += 1
|
||
}
|
||
i -= 1
|
||
j -= 1
|
||
|
||
// 向左
|
||
while (j >= 0 && res(i)(j) == 0) {
|
||
res(i)(j) = num
|
||
num += 1
|
||
j -= 1
|
||
}
|
||
i -= 1
|
||
j += 1
|
||
|
||
// 向上
|
||
while (i >= 0 && res(i)(j) == 0) {
|
||
res(i)(j) = num
|
||
num += 1
|
||
i -= 1
|
||
}
|
||
i += 1
|
||
j += 1
|
||
}
|
||
res
|
||
}
|
||
}
|
||
```
|
||
### C#：
|
||
|
||
```csharp
|
||
public class Solution {
|
||
public int[][] GenerateMatrix(int n) {
|
||
int[][] answer = new int[n][];
|
||
for(int i = 0; i < n; i++)
|
||
answer[i] = new int[n];
|
||
int start = 0;
|
||
int end = n - 1;
|
||
int tmp = 1;
|
||
while(tmp < n * n)
|
||
{
|
||
for(int i = start; i < end; i++) answer[start][i] = tmp++;
|
||
for(int i = start; i < end; i++) answer[i][end] = tmp++;
|
||
for(int i = end; i > start; i--) answer[end][i] = tmp++;
|
||
for(int i = end; i > start; i--) answer[i][start] = tmp++;
|
||
start++;
|
||
end--;
|
||
}
|
||
if(n % 2 == 1) answer[n / 2][n / 2] = tmp;
|
||
return answer;
|
||
}
|
||
}
|
||
```
|
||
|
||
### Ruby:
|
||
```ruby
|
||
def generate_matrix(n)
|
||
result = Array.new(n) { Array.new(n, 0) }
|
||
#循环次数
|
||
loop_times = 0
|
||
#步长
|
||
step = n - 1
|
||
val = 1
|
||
|
||
|
||
while loop_times < n / 2
|
||
#模拟从左向右
|
||
for i in 0..step - 1
|
||
#行数不变，列数变
|
||
result[loop_times][i+loop_times] = val
|
||
val += 1
|
||
end
|
||
|
||
#模拟从上到下
|
||
for i in 0..step - 1
|
||
#列数不变，行数变
|
||
result[i+loop_times][n-loop_times-1] = val
|
||
val += 1
|
||
end
|
||
|
||
#模拟从右到左
|
||
for i in 0..step - 1
|
||
#行数不变，列数变
|
||
result[n-loop_times-1][n-loop_times-i-1] = val
|
||
val += 1
|
||
end
|
||
|
||
#模拟从下到上
|
||
for i in 0..step - 1
|
||
#列数不变，行数变
|
||
result[n-loop_times-i-1][loop_times] = val
|
||
val += 1
|
||
end
|
||
|
||
loop_times += 1
|
||
step -= 2
|
||
end
|
||
|
||
#如果是奇数，则填充最后一个元素
|
||
result[n/2][n/2] = n**2 if n % 2
|
||
|
||
return result
|
||
|
||
end
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|