LeetCode48_Rotate Image

Rotate Image

问题描述

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]]

简单分析

这道题就是把给定的矩阵进行顺时针90度旋转。只是简单的数组操作，但是也有两种解法。

一种解法是直接对数组进行90度旋转赋值，代码如下。

代码

public void rotate(int[][] matrix) {        int length = matrix.length;        int iMax = length>>1;        for(int n=0;n<=iMax;n++){            int l = length-(n<<1);            for(int i=0;i<l-1;i++){                //将第一行的数据保留下来                int temp=matrix[n][length-1-n-i];                //将左边旋转移到上边                matrix[n][length-1-n-i]=matrix[i+n][n];                //将底边转移到左边                matrix[i+n][n]=matrix[length-1-n][i+n];                //将右边转移到底边                matrix[length-1-n][i+n]=matrix[length-1-n-i][matrix.length-1-n];                //将上边的值转移到右边                matrix[length-1-n-i][length-1-n]=temp;            }         }    }

另外一种是LeetCode中排名靠前的一种方法，

1. 首先对矩阵按照对角线进行镜像处理，
2. 然后在按照中轴线进行翻转。

这样得到的也是旋转90度的结果。
读者可以自己实现一下代码，感觉两种算法没有优劣之分，都差不多。

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GitHub地址https://github.com/yanqinghe/leetcode