自己的思路：从外到内一圈圈顺时针旋转90度，坐标映射问题。
Leetcode讨论区有很多有趣巧妙的思路，列举两个点赞率较高思路：
1）上下颠倒，然后转置
/*
* clockwise rotate
* first reverse up to down, then swap the symmetry
* 1 2 3 7 8 9 7 4 1
* 4 5 6 => 4 5 6 => 8 5 2
* 7 8 9 1 2 3 9 6 3
*/
2）转置，然后左右颠倒
若需要逆时针，则分别是：
1）左右颠倒，然后转置
2）转置，然后上下颠倒
[ref]
https://leetcode.com/discuss/20589/a-common-method-to-rotate-the-image
public class Solution {
public void rotate1(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return;
int rowBeg = 0, rowEnd = matrix.length - 1;
int colBeg = 0, colEnd = matrix[0].length - 1;
while (rowBeg <= rowEnd) {
int offset = colEnd - colBeg - 1;
for (int j = 0; j <= offset; j++) {
int save = matrix[rowBeg][colBeg + j];
matrix[rowBeg][colBeg + j] = matrix[rowEnd - j][colBeg];
matrix[rowEnd - j][colBeg] = matrix[rowEnd][colEnd - j];
matrix[rowEnd][colEnd - j] = matrix[rowBeg + j][colEnd];
matrix[rowBeg + j][colEnd] = save;
}
rowBeg++; rowEnd--;
colBeg++; colEnd--;
}
}
// first transpose, then flip the matrix horizontally
public void rotate(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return;
int rows = matrix.length, cols = matrix[0].length;
for (int i = 0; i < rows; i++) {
for (int j = i + 1; j < cols; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = tmp;
}
}
int half = cols / 2;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < half; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[i][cols -1 - j];
matrix[i][cols -1 - j] = tmp;
}
}
}
}