题目描述

原题链接

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：

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：

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]]

提示：

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

个人解法

Javascript

方法一：使用辅助数组

/*
 * @lc app=leetcode.cn id=48 lang=javascript
 *
 * [48] 旋转图像
 */
// @lc code=start
/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
 var rotate = function (matrix) {
  const len = matrix.length;
  const myMatrix = new Array(len);
  for (let i = 0; i < len; i++) {
    myMatrix[i] = [];
    for (let j = 0; j < len; j++) {
      myMatrix[i].push(matrix[j][i]);
    }
  }
  for (let i = 0; i < len; i++) {
    for (let j = 0; j < len; j++) {
      matrix[i][j] = myMatrix[i][len - j - 1];
    }
  }
};
// @lc code=end

Java

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        int[][] matrix_new = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                matrix_new[j][n - i - 1] = matrix[i][j];
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                matrix[i][j] = matrix_new[i][j];
            }
        }
    }
}

其他解法

Java

原地旋转

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n / 2; ++i) {
            for (int j = 0; j < (n + 1) / 2; ++j) {
                int 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;
            }
        }
    }
}

用翻转代替旋转

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        // 水平翻转
        for (int i = 0; i < n / 2; ++i) {
            for (int j = 0; j < n; ++j) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[n - i - 1][j];
                matrix[n - i - 1][j] = temp;
            }
        }
        // 主对角线翻转
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
    }
}

Javascript

原地旋转

var rotate = function(matrix) {
    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;
        }
    }
};

方法三：用翻转代替旋转

var rotate = function(matrix) {
    const n = matrix.length;
    // 水平翻转
    for (let i = 0; i < Math.floor(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]];
        }
    }
    // 主对角线翻转
    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]];
        }
    }
};