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

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

一刷
題解：
坐標(biāo)的變換為：
(i, j)->(j, n-1-i)->(n-1-i, n-1-j)->(n-1-j, i)->(i, j)

此時(shí)，注意i和j的范圍
當(dāng)n為偶數(shù)時(shí)，i,j的范圍相同;
當(dāng)n為奇數(shù)時(shí)，i比j最大值大1,則中線上的值只變動(dòng)一次，避免產(chǎn)生沖突。（或者j比i最大值大1）

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