leetcode48 Rotate Image 旋转图片

题目：

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]
]

主要意思就是将数组顺时针旋转90度，要求是不能另开一个数组空间来进行操作。

解法思想：建立一个temp数组，对于同一层的旋转，先用temp保存一条边的值，之后进行旋转，最后将temp返回旋转后的最后一条边。整个过程由最外层向里进行。

代码：

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