Description
Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.
Example 1:
Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]
Example 2:
Input: n = 1
Output: [[1]]
Constraints:
Solutions
Solution 1: Simulation
We can directly simulate the process of generating the spiral matrix.
Define a 2D array ans to store the spiral matrix. Use i and j to represent the current row and column indices, and use k to represent the current direction index. dirs represents the mapping between direction indices and directions.
Starting from 1, fill each position in the matrix sequentially. After filling a position, calculate the row and column indices of the next position. If the next position is out of bounds or has already been filled, change the direction and then calculate the row and column indices of the next position.
The time complexity is O(n^2), where n is the side length of the matrix. Ignoring the space consumption of the answer array, the space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScript
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
ans = [[0] * n for _ in range(n)]
dirs = (0, 1, 0, -1, 0)
i = j = k = 0
for v in range(1, n * n + 1):
ans[i][j] = v
x, y = i + dirs[k], j + dirs[k + 1]
if x < 0 or x >= n or y < 0 or y >= n or ans[x][y]:
k = (k + 1) % 4
i, j = i + dirs[k], j + dirs[k + 1]
return ans(code-box)
class Solution {
public int[][] generateMatrix(int n) {
int[][] ans = new int[n][n];
final int[] dirs = {0, 1, 0, -1, 0};
int i = 0, j = 0, k = 0;
for (int v = 1; v <= n * n; ++v) {
ans[i][j] = v;
int x = i + dirs[k], y = j + dirs[k + 1];
if (x < 0 || x >= n || y < 0 || y >= n || ans[x][y] != 0) {
k = (k + 1) % 4;
}
i += dirs[k];
j += dirs[k + 1];
}
return ans;
}
}(code-box)
class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> ans(n, vector<int>(n, 0));
const int dirs[5] = {0, 1, 0, -1, 0};
int i = 0, j = 0, k = 0;
for (int v = 1; v <= n * n; ++v) {
ans[i][j] = v;
int x = i + dirs[k], y = j + dirs[k + 1];
if (x < 0 || x >= n || y < 0 || y >= n || ans[x][y] != 0) {
k = (k + 1) % 4;
}
i += dirs[k];
j += dirs[k + 1];
}
return ans;
}
};(code-box)
func generateMatrix(n int) [][]int {
ans := make([][]int, n)
for i := range ans {
ans[i] = make([]int, n)
}
dirs := [5]int{0, 1, 0, -1, 0}
i, j, k := 0, 0, 0
for v := 1; v <= n*n; v++ {
ans[i][j] = v
x, y := i+dirs[k], j+dirs[k+1]
if x < 0 || x >= n || y < 0 || y >= n || ans[x][y] != 0 {
k = (k + 1) % 4
}
i += dirs[k]
j += dirs[k+1]
}
return ans
}(code-box)
function generateMatrix(n: number): number[][] {
const ans: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
const dirs = [0, 1, 0, -1, 0];
let [i, j, k] = [0, 0, 0];
for (let v = 1; v <= n * n; v++) {
ans[i][j] = v;
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (x < 0 || x >= n || y < 0 || y >= n || ans[x][y] !== 0) {
k = (k + 1) % 4;
}
i += dirs[k];
j += dirs[k + 1];
}
return ans;
}(code-box)
impl Solution {
pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {
let mut ans = vec![vec![0; n as usize]; n as usize];
let dirs = [0, 1, 0, -1, 0];
let (mut i, mut j, mut k) = (0, 0, 0);
for v in 1..=n * n {
ans[i as usize][j as usize] = v;
let (x, y) = (i + dirs[k], j + dirs[k + 1]);
if x < 0 || x >= n || y < 0 || y >= n || ans[x as usize][y as usize] != 0 {
k = (k + 1) % 4;
}
i += dirs[k];
j += dirs[k + 1];
}
ans
}
}(code-box)
/**
* @param {number} n
* @return {number[][]}
*/
var generateMatrix = function (n) {
const ans = Array.from({ length: n }, () => Array(n).fill(0));
const dirs = [0, 1, 0, -1, 0];
let [i, j, k] = [0, 0, 0];
for (let v = 1; v <= n * n; v++) {
ans[i][j] = v;
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (x < 0 || x >= n || y < 0 || y >= n || ans[x][y] !== 0) {
k = (k + 1) % 4;
}
i += dirs[k];
j += dirs[k + 1];
}
return ans;
};(code-box)
