[LeetCode] 48. Rotate Image 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.

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]
]

一个n x n的二维矩阵表示一个图像，将图像顺时针旋转90度。要求in-place，所以就不能用额外的空间了。

解法1: 先以对角线为轴翻转得到其转置矩阵，再以中间竖轴翻转。

1  2  3　　　 　1  4  7　　　　 7  4  1

4  5  6　-->　  2  5  8　 -->   8  5  2　　

7  8  9 　　　   3  6  9　　   　9  6  3

解法2: 先以反对角线翻转，在以中间水平轴翻转。

1  2  3　　　 　9  6  3　　　　7  4  1

4  5  6　-->　 8  5  2　 -->   8  5  2　　

7  8  9 　　　  7  4  1　　　　 9  6  3

 

Java：

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public class Solution {     public void rotate(int[][] matrix) {         int n = matrix.length;                   // along the left top to right bottom diagonal line, swap symmetrical pair         for(int i=0; i<n; i++) {  // for each row             for(int j=i+1; j<n; j++) {  // for each number                 // swap the pair                 int temp = matrix[i][j];                 matrix[i][j] = matrix[j][i];                 matrix[j][i] = temp;             }         }                   // flip each row horizontally         for(int i=0; i<n; i++) {             for(int j=0; j<n/2; j++) {                 int temp = matrix[i][j];                 matrix[i][j] = matrix[i][n-1-j];                 matrix[i][n-1-j] = temp;                               }         }     } }

Python: T: O(n^2), S: O(1)

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class Solution:     # @param matrix, a list of lists of integers     # @return a list of lists of integers     def rotate(self, matrix):         n = len(matrix)                   # anti-diagonal mirror         for i in xrange(n):             for j in xrange(n - i):                 matrix[i][j], matrix[n-1-j][n-1-i] = matrix[n-1-j][n-1-i], matrix[i][j]                   # horizontal mirror         for i in xrange(n / 2):             for j in xrange(n):                 matrix[i][j], matrix[n-1-i][j] = matrix[n-1-i][j], matrix[i][j]                           return matrix

C++:

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class Solution { public:     void rotate(vector<vector<int> > &matrix) {         int n = matrix.size();         for (int i = 0; i < n; ++i) {             for (int j = i + 1; j < n; ++j) {                 swap(matrix[i][j], matrix[j][i]);             }             reverse(matrix[i].begin(), matrix[i].end());         }     } };

　　

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