48. Rotate Image

难度: Medium

题目描述

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

 

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

 

Constraints:

• n == matrix.length == matrix[i].length
• 1 <= n <= 20
• -1000 <= matrix[i][j] <= 1000

解题思路

代码实现

解决方案

java

class Solution {
    public void rotate(int[][] matrix) {
        verticalMirrorInPlace(matrix);
        transposeInPlace(matrix);

    }

    public static void transposeInPlace(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return;
        }
        int n = matrix.length;
        // 遍历上三角区域（i < j），交换 matrix[i][j] 和 matrix[j][i]
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) { // j 从 i+1 开始，避免重复交换
                // 交换元素
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
    }

    public static void verticalMirrorInPlace(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return;
        }
        int n = matrix.length;
        // 遍历前 n/2 行，与对称行交换
        for (int i = 0; i < n / 2; i++) {
            int mirrorRow = n - 1 - i; // 对称行索引
            // 逐元素交换第 i 行和第 mirrorRow 行
            for (int j = 0; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[mirrorRow][j];
                matrix[mirrorRow][j] = temp;
            }
        }
    }

    // 原地水平镜像（左右翻转）n×n 方阵
    public static void horizontalMirrorInPlace(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return;
        }
        int n = matrix.length;
        // 遍历每一行
        for (int i = 0; i < n; i++) {
            // 遍历每行的前 n/2 个元素，与对称位置交换
            for (int j = 0; j < n / 2; j++) {
                int mirrorCol = n - 1 - j; // 对称列索引
                // 交换第 i 行的第 j 个和第 mirrorCol 个元素
                int temp = matrix[i][j];
                matrix[i][j] = matrix[i][mirrorCol];
                matrix[i][mirrorCol] = temp;
            }
        }
    }
}