LeetCode-----48. Rotate Image(二维矩阵旋转90度)

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?

坐标旋转(顺时针)：

0°/360°：arr[i][j]

90°（相当于逆时针270°）：a[i][j]=b[j][n-i-1];

180°：a[i][j] == b[n - i - 1][n - j -1]

270°：a[i][j] == b[n - j - 1][i]

//转置矩阵  行变列

创建一个新的矩阵：b[i][j] = a[j][i];

思路一：

对原矩阵进行求转置矩阵，转置矩阵每一行进行反转即可得到旋转90°后的矩阵

思路二：cmp

public class Solution {
public void rotate(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return;
}

int length = matrix.length;

for (int i = 0; i < length / 2; i++) {
for (int j = 0; j < (length + 1) / 2; j++){
int tmp = matrix[i][j];
matrix[i][j] = matrix[length - j - 1][i];
matrix[length -j - 1][i] = matrix[length - i - 1][length - j - 1];
matrix[length - i - 1][length - j - 1] = matrix[j][length - i - 1];
matrix[j][length - i - 1] = tmp;
}
}
}
}

判断AB矩阵旋转角度：

import java.util.Scanner;
public class SwitchMartix{
private static Scanner sc=new Scanner(System.in);
private static int n; //方阵规模 //产生n阶方阵 public static int[][] getMatrix(){
int[][]A=new int

;
for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ A[i][j]=sc.nextInt(); } } return A; } //获取旋转角度 angle: -1表示不是旋转矩阵；旋转矩阵角度——0 90 180 270
public static int getAngle(int[][]a,int[][]b){ int angle = 0;
for(int i = 0; i < n; i ++){
for(int j = 0; j < n; j ++){
//开始遍历
if(angle == 0){ //假设修正法 角度依次增加 if(a[i][j] == b[i][j]){
continue; }else{

angle = 90;                    }                }

if(angle == 90){  //假设修正法

if(a[i][j] == b[j][n - i - 1]){
continue;                    }else{

angle = 180;                    }                }

if(angle == 180){  //假设修正法                   if(a[i][j] == b[n - i - 1][n - j -1]){                        continue;                    }else{                         angle = 270;                    }                }                     if(angle == 270){
//假设修正法
if(a[i][j] == b[n - j - 1][i]){
continue;                    }else{

return -1; //不是旋转矩阵，直接返回                  }                }                          }//for
}//for
return angle;    }