给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。
示例 1：
输入：n = 3 输出：[[1,2,3],[8,9,4],[7,6,5]]
示例 2：
输入：n = 1 输出：[[1]]
提示：
1 <= n <= 20
方法一：模拟
直接模拟螺旋矩阵的生成过程。
定义一个二维数组 ans，用于存储螺旋矩阵。用 i 和 j 分别表示当前位置的行号和列号，用 k 表示当前的方向编号，dirs 表示方向编号与方向的对应关系。
从 1 开始，依次填入矩阵中的每个位置。每次填入一个位置后，计算下一个位置的行号和列号，如果下一个位置不在矩阵中或者已经被填过，则改变方向，再计算下一个位置的行号和列号。
时间复杂度
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
ans = [[0] * n for _ in range(n)]
dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
i = j = k = 0
for v in range(1, n * n + 1):
ans[i][j] = v
x, y = i + dirs[k][0], j + dirs[k][1]
if x < 0 or y < 0 or x >= n or y >= n or ans[x][y]:
k = (k + 1) % 4
x, y = i + dirs[k][0], j + dirs[k][1]
i, j = x, y
return ansclass Solution {
public int[][] generateMatrix(int n) {
int[][] ans = new int[n][n];
int i = 0, j = 0, k = 0;
int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
for (int v = 1; v <= n * n; ++v) {
ans[i][j] = v;
int x = i + dirs[k][0], y = j + dirs[k][1];
if (x < 0 || y < 0 || x >= n || y >= n || ans[x][y] > 0) {
k = (k + 1) % 4;
x = i + dirs[k][0];
y = j + dirs[k][1];
}
i = x;
j = y;
}
return ans;
}
}class Solution {
public:
const int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> ans(n, vector<int>(n));
int i = 0, j = 0, k = 0;
for (int v = 1; v <= n * n; ++v) {
ans[i][j] = v;
int x = i + dirs[k][0], y = j + dirs[k][1];
if (x < 0 || y < 0 || x >= n || y >= n || ans[x][y]) {
k = (k + 1) % 4;
x = i + dirs[k][0], y = j + dirs[k][1];
}
i = x, j = y;
}
return ans;
}
};func generateMatrix(n int) [][]int {
ans := make([][]int, n)
for i := range ans {
ans[i] = make([]int, n)
}
dirs := [4][2]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}}
var i, j, k int
for v := 1; v <= n*n; v++ {
ans[i][j] = v
x, y := i+dirs[k][0], j+dirs[k][1]
if x < 0 || y < 0 || x >= n || y >= n || ans[x][y] > 0 {
k = (k + 1) % 4
x, y = i+dirs[k][0], j+dirs[k][1]
}
i, j = x, y
}
return ans
}/**
* @param {number} n
* @return {number[][]}
*/
var generateMatrix = function (n) {
const ans = new Array(n).fill(0).map(() => new Array(n).fill(0));
let [i, j, k] = [0, 0, 0];
const dirs = [
[0, 1],
[1, 0],
[0, -1],
[-1, 0],
];
for (let v = 1; v <= n * n; ++v) {
ans[i][j] = v;
let [x, y] = [i + dirs[k][0], j + dirs[k][1]];
if (x < 0 || y < 0 || x >= n || y >= n || ans[x][y] > 0) {
k = (k + 1) % 4;
[x, y] = [i + dirs[k][0], j + dirs[k][1]];
}
[i, j] = [x, y];
}
return ans;
};function generateMatrix(n: number): number[][] {
let ans = Array.from({ length: n }, v => new Array(n));
let dir = [
[0, 1],
[1, 0],
[0, -1],
[-1, 0],
];
let i = 0,
j = 0;
for (let cnt = 1, k = 0; cnt <= n * n; cnt++) {
ans[i][j] = cnt;
let x = i + dir[k][0],
y = j + dir[k][1];
if (x < 0 || x == n || y < 0 || y == n || ans[x][y]) {
k = (k + 1) % 4;
(x = i + dir[k][0]), (y = j + dir[k][1]);
}
(i = x), (j = y);
}
return ans;
}function generateMatrix(n: number): number[][] {
const res = new Array(n).fill(0).map(() => new Array(n).fill(0));
let num = 1;
for (let i = 0; i < Math.floor(n / 2); i++) {
for (let j = i; j < n - i - 1; j++) {
res[i][j] = num++;
}
for (let j = i; j < n - i - 1; j++) {
res[j][n - i - 1] = num++;
}
for (let j = i; j < n - i - 1; j++) {
res[n - i - 1][n - j - 1] = num++;
}
for (let j = i; j < n - i - 1; j++) {
res[n - j - 1][i] = num++;
}
}
if (n % 2 === 1) {
res[n >> 1][n >> 1] = num;
}
return res;
}impl Solution {
pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {
let n = n as usize;
let mut res = vec![vec![0; n]; n];
let mut num = 1;
for i in 0..n / 2 {
for j in i..n - i - 1 {
res[i][j] = num;
num += 1;
}
for j in i..n - i - 1 {
res[j][n - i - 1] = num;
num += 1;
}
for j in i..n - i - 1 {
res[n - i - 1][n - j - 1] = num;
num += 1;
}
for j in i..n - i - 1 {
res[n - j - 1][i] = num;
num += 1;
}
}
if n % 2 == 1 {
res[n >> 1][n >> 1] = num;
}
res
}
}