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2022.03.25-第60题-48. 旋转图像 #62

Open yanggengzhen123 opened 2 years ago

yanggengzhen123 commented 2 years ago

https://leetcode-cn.com/problems/rotate-image/ 给定一个 n × n 的二维矩阵 matrix 表示一个图像。请你将图像顺时针旋转 90 度。 你必须在 原地 旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。 示例 1: image 输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[[7,4,1],[8,5,2],[9,6,3]]

示例 2: image 输入:matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] 输出:[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

icodeish commented 2 years ago
var rotate = function (matrix) {
    let len = matrix.length
    let new_matrix = new Array(len).fill(0).map(() => new Array(len).fill(0))
    for (let i = 0; i < len; i++) {
        for (let j = 0; j < len; j++) {
            //new_matrix[j][len -1-i] = matrix[i][j]
            new_matrix[i][j] = matrix[len - 1 - j][i]
        }
    }
    for (let i = 0; i < len; i++) {
        for (let j = 0; j < len; j++) {
            matrix[i][j] = new_matrix[i][j]
        }
    }
    return matrix
}
BambooSword commented 2 years ago
function rotate(matrix: number[][]): void {
    // 矩阵的旋转
    const n = matrix.length;
    for (let i = 0; i < Math.floor(n / 2); ++i) {
        for (let j = 0; j < Math.floor((n + 1) / 2); ++j) {
            const temp = matrix[i][j];
            matrix[i][j] = matrix[n - j - 1][i];
            matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
            matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
            matrix[j][n - i - 1] = temp;
        }
    }
};
yanggengzhen123 commented 2 years ago
*
function rotate(matrix: number[][]): void {
    const n = matrix.length;
    // 水平翻转
    for(let i = 0; i < n / 2; i++){
        for(let j = 0; j < n; j++){
            [matrix[i][j], matrix[n - i - 1][j]] = [matrix[n - i - 1][j], matrix[i][j]]
        }
    }
    // 对角线翻转 (左上到右下为中心线)
    for(let i = 0; i < n; i++){
        for(let j = 0; j < i; j++){
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]]
        }
    }
}