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[n][n];  
  for(int i=0;i

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;    }


你可能感兴趣的:(矩阵操作,LeetCode)