0048. Rotate Image (M)

佚名 4年前 (2020-07-01) 算法 691人围观抢沙发百度已收录

Rotate Image (M)

题目

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

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Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOTallocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

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

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

题意

将给定的矩阵顺时针旋转90°，要求只能在原矩阵上进行修改。

思路

比较直接的方法是模拟旋转，如下图，经过旋转后，第一列变为第一行，第二列变为第二行……而第一行变为最后一列，第二行变为最后第二列……由此可以得到旋转规律：\((i,\ j) \rightarrow (j,\ n-i-1) \rightarrow (n-i-1,\ n-j-1) \rightarrow (n-j-1,\ i) \rightarrow (i,\ j)\)。

为了不重复操作，只将下图中的区域①旋转3次得到区域②、③、④。

另一种比较取巧的方法是，如下图，先沿着中心水平线上下翻转，再沿着左上到右下的主对角线翻转，就完成了顺时针旋转90°。

代码实现

Java

旋转法

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

翻转法

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

    private void swap(int[][] matrix, int i, int j, int x, int y) {
        int temp = matrix[i][j];
        matrix[i][j] = matrix[x][y];
        matrix[x][y] = temp;
    }
}

JavaScript

旋转法

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function (matrix) {
  let n = matrix.length
  for (let i = 0; i < Math.trunc(n / 2); i++) {
    for (let j = i; j <= n - 1 - i; j++) {
      let tmp = matrix[i][j]
      matrix[i][j] = matrix[n - 1 - j][i]
      matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]
      matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]
      matrix[j][n - 1 - i] = tmp
    }
  }
}

翻转法

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function (matrix) {
  let n = matrix.length

  for (let i = 0; i < Math.trunc(n / 2); i++) {
    for (let j = 0; j < n; j++) {
      [matrix[i][j], matrix[n - 1 - i][j]] = [matrix[n - 1 - i][j], matrix[i][j]]
    }
  }

  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]]
    }
  }
}