螺旋矩阵 II

59. 螺旋矩阵 II

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

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螺旋矩阵

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上次编辑到这里，代码来自缓存点击恢复默认模板

class Solution { public: vector<vector<int>> generateMatrix(int n) { } };

python3 解法, 执行用时: 46 ms, 内存消耗: 16.4 MB, 提交时间: 2024-04-20 16:23:47

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        matrix = [[0] * n for _ in range(n)]
        num = 1
        left, right, top, bottom = 0, n - 1, 0, n - 1

        while left <= right and top <= bottom:
            for col in range(left, right + 1):
                matrix[top][col] = num
                num += 1
            for row in range(top + 1, bottom + 1):
                matrix[row][right] = num
                num += 1
            if left < right and top < bottom:
                for col in range(right - 1, left, -1):
                    matrix[bottom][col] = num
                    num += 1
                for row in range(bottom, top, -1):
                    matrix[row][left] = num
                    num += 1
            left += 1
            right -= 1
            top += 1
            bottom -= 1

        return matrix

javascript 解法, 执行用时: 56 ms, 内存消耗: 49.7 MB, 提交时间: 2024-04-20 16:23:20

/**
 * @param {number} n
 * @return {number[][]}
 */
var generateMatrix = function(n) {
    const maxNum = n * n;
    let curNum = 1;
    const matrix = new Array(n).fill(0).map(() => new Array(n).fill(0));
    let row = 0, column = 0;
    const directions = [[0, 1], [1, 0], [0, -1], [-1, 0]]; // 右下左上
    let directionIndex = 0;
    while (curNum <= maxNum) {
        matrix[row][column] = curNum;
        curNum++;
        const nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
        if (nextRow < 0 || nextRow >= n || nextColumn < 0 || nextColumn >= n || matrix[nextRow][nextColumn] !== 0) {
            directionIndex = (directionIndex + 1) % 4; // 顺时针旋转至下一个方向
        }
        row = row + directions[directionIndex][0];
        column = column + directions[directionIndex][1];
    }
    return matrix;
};

javascript 解法, 执行用时: 55 ms, 内存消耗: 49.2 MB, 提交时间: 2024-04-20 16:22:37

/**
 * @param {number} n
 * @return {number[][]}
 */
var generateMatrix = function(n) {
    let num = 1;
    const matrix = new Array(n).fill(0).map(() => new Array(n).fill(0));
    let left = 0, right = n - 1, top = 0, bottom = n - 1;
    while (left <= right && top <= bottom) {
        for (let column = left; column <= right; column++) {
            matrix[top][column] = num;
            num++;
        }
        for (let row = top + 1; row <= bottom; row++) {
            matrix[row][right] = num;
            num++;
        }
        if (left < right && top < bottom) {
            for (let column = right - 1; column > left; column--) {
                matrix[bottom][column] = num;
                num++;
            }
            for (let row = bottom; row > top; row--) {
                matrix[row][left] = num;
                num++;
            }
        }
        left++;
        right--;
        top++;
        bottom--;
    }
    return matrix;
};

golang 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2024-04-20 16:22:21

func generateMatrix(n int) [][]int {
    matrix := make([][]int, n)
    for i := range matrix {
        matrix[i] = make([]int, n)
    }
    num := 1
    left, right, top, bottom := 0, n-1, 0, n-1
    for left <= right && top <= bottom {
        for column := left; column <= right; column++ {
            matrix[top][column] = num
            num++
        }
        for row := top + 1; row <= bottom; row++ {
            matrix[row][right] = num
            num++
        }
        if left < right && top < bottom {
            for column := right - 1; column > left; column-- {
                matrix[bottom][column] = num
                num++
            }
            for row := bottom; row > top; row-- {
                matrix[row][left] = num
                num++
            }
        }
        left++
        right--
        top++
        bottom--
    }
    return matrix
}

golang 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2024-04-20 16:22:09

type pair struct{ x, y int }
var dirs = []pair{{0, 1}, {1, 0}, {0, -1}, {-1, 0}} // 右下左上

func generateMatrix(n int) [][]int {
    matrix := make([][]int, n)
    for i := range matrix {
        matrix[i] = make([]int, n)
    }
    row, col, dirIdx := 0, 0, 0
    for i := 1; i <= n*n; i++ {
        matrix[row][col] = i
        dir := dirs[dirIdx]
        if r, c := row+dir.x, col+dir.y; r < 0 || r >= n || c < 0 || c >= n || matrix[r][c] > 0 {
            dirIdx = (dirIdx + 1) % 4 // 顺时针旋转至下一个方向
            dir = dirs[dirIdx]
        }
        row += dir.x
        col += dir.y
    }
    return matrix
}

cpp 解法, 执行用时: 0 ms, 内存消耗: 7.6 MB, 提交时间: 2024-04-20 16:21:43

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        int num = 1;
        vector<vector<int>> matrix(n, vector<int>(n));
        int left = 0, right = n - 1, top = 0, bottom = n - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                matrix[top][column] = num;
                num++;
            }
            for (int row = top + 1; row <= bottom; row++) {
                matrix[row][right] = num;
                num++;
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    matrix[bottom][column] = num;
                    num++;
                }
                for (int row = bottom; row > top; row--) {
                    matrix[row][left] = num;
                    num++;
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return matrix;
    }
};

java 解法, 执行用时: 0 ms, 内存消耗: 40.6 MB, 提交时间: 2024-04-20 16:21:29

class Solution {
    public int[][] generateMatrix(int n) {
        int num = 1;
        int[][] matrix = new int[n][n];
        int left = 0, right = n - 1, top = 0, bottom = n - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                matrix[top][column] = num;
                num++;
            }
            for (int row = top + 1; row <= bottom; row++) {
                matrix[row][right] = num;
                num++;
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    matrix[bottom][column] = num;
                    num++;
                }
                for (int row = bottom; row > top; row--) {
                    matrix[row][left] = num;
                    num++;
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return matrix;
    }
}

java 解法, 执行用时: 0 ms, 内存消耗: 40.3 MB, 提交时间: 2024-04-20 16:21:16

class Solution {
    public int[][] generateMatrix(int n) {
        int maxNum = n * n;
        int curNum = 1;
        int[][] matrix = new int[n][n];
        int row = 0, column = 0;
        int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // 右下左上
        int directionIndex = 0;
        while (curNum <= maxNum) {
            matrix[row][column] = curNum;
            curNum++;
            int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
            if (nextRow < 0 || nextRow >= n || nextColumn < 0 || nextColumn >= n || matrix[nextRow][nextColumn] != 0) {
                directionIndex = (directionIndex + 1) % 4; // 顺时针旋转至下一个方向
            }
            row = row + directions[directionIndex][0];
            column = column + directions[directionIndex][1];
        }
        return matrix;
    }
}

python3 解法, 执行用时: 40 ms, 内存消耗: 15 MB, 提交时间: 2022-06-21 11:53:59

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        matrix = [[0] * n for _ in range(n)]
        row, col, dirIdx = 0, 0, 0
        for i in range(n * n):
            matrix[row][col] = i + 1
            dx, dy = dirs[dirIdx]
            r, c = row + dx, col + dy
            if r < 0 or r >= n or c < 0 or c >= n or matrix[r][c] > 0:
                dirIdx = (dirIdx + 1) % 4   # 顺时针旋转至下一个方向
                dx, dy = dirs[dirIdx]
            row, col = row + dx, col + dy
        return matrix

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