leetcode——[048]Rotate Image旋转图像

题目

给定一个 n × n 的二维矩阵表示一个图像。

将图像顺时针旋转 90 度。

说明：

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。

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

Rotate the image by 90 degrees (clockwise).

Note:

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.

示例 1:

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

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

示例 2:

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

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

解题方法

拆分为翻折

将矩阵顺时针旋转90°，分解为绕轴翻转的步骤。也就是先绕对角线翻转，然后绕两腰中位线翻转。要求直接操作原矩阵，所以不能直接使用新矩阵来保存结果，使用一个变量来保存被替换的单元。这段代码跑了2ms，超过93.40%的java提交。

public class Solution {
    // Method_01  2ms 93.40%
    public void rotate(int[][] matrix) {
        int tmp;
        int length = matrix.length;
        // fold along the diagonal
        for (int i = 0; i < length; i++) {
            for (int j = 0; j < length - i; j++) {
                tmp = matrix[length - j - 1][length - i - 1];
                matrix[length - j - 1][length - i - 1] = matrix[i][j];
                matrix[i][j] = tmp;
            }
        }
        // fold along the median line
        for (int i = 0; i < length / 2; i++) {
            for (int j = 0; j < length; j++) {
                tmp = matrix[length - i - 1][j];
                matrix[length - i - 1][j] = matrix[i][j];
                matrix[i][j] = tmp;
            }
        }
    }
}

旋转对应坐标变换

顺时针旋转90°，对应360°/90°=4个单元进行逐一替换，对应单元的坐标替换关系为:

[i, j] -> [j, matrix.length - 1 - i];

[j, matrix.length - 1 - i] -> [matrix.length - 1- i, matrix.length - 1 - j];

[matrix.length - 1- i, matrix.length - 1 - j] -> [matrix.length - 1- j, i];

[matrix.length - 1- j, i] -> [i, j]。

按这种坐标转换，遍历矩阵的四分之一即可。这段代码也跑了2ms，超过了93.40%的java提交。

class Solution {
    // Method_02  2ms  93.40%
    public void rotate(int[][] matrix) {
        int tmp;
        int length = matrix.length;
        for (int i = 0; i < length / 2; i++) {
            for (int j = 0; j < (length + 1) / 2; j++) {
                tmp = matrix[i][j];
                int i1 = j;
                int j1 = matrix.length - 1 - i;
                int i2 = j1;
                int j2 = matrix.length - 1 - j;
                int i3 = j2;
                int j3 = i;
                matrix[i][j] = matrix[i3][j3];
                matrix[i3][j3] = matrix[i2][j2];
                matrix[i2][j2] = matrix[i1][j1];
                matrix[i1][j1] = tmp;
            }
        }
    }
}

结语

Keep coding！