【LeetCode】解题48：Rotate Image

最新推荐文章于 2020-12-02 21:02:12 发布

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最新推荐文章于 2020-12-02 21:02:12 发布

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LeetCode解题 48：Rotate Image

Problem 48: Rotate Image [Medium]
解题思路
Solution (Java)

Problem 48: Rotate Image [Medium]

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]
]

来源：LeetCode

解题思路

将数组看作一个由N/2个正方形圈组成的套圈，每一次循环都将一个圈在原地顺时针旋转90°，能够保证不会重复旋转，并且不需要额外数组空间。
如下图所示：
在这里插入图片描述
一次旋转圈1：

第二次旋转圈2：

具体过程：

循环N/2次，每次旋转的正方形的四个顶点为：
$top\_left = [side, side]，$
$top\_right = [side, N-side-1]，$
$bottom\_right = [N-side-1, N-side-1]，$
$bottom\_left = [N-side-1, side]$
旋转正方形时需要调换 $4 * (N - 2 * s i d e - 1)$ 个点的位置。每四个点为一组，内部进行旋转替换，循环该过程 $(N - 2 * s i d e - 1)$ 次。
分组如图所示：

运行结果：
在这里插入图片描述

Solution (Java)

class Solution {
    public void rotate(int[][] matrix) {
        int N = matrix.length;
        if(N == 0 || N == 1) return;
        for(int i = 0; i < N / 2; i++){
            rotate_sides(matrix, i, N);
        }
    }
    private void rotate_sides(int[][] matrix, int side, int N){
        for(int i = side; i < N - side - 1; i++){
            int top = matrix[side][i];
            int right = matrix[i][N-side-1];
            int bottom = matrix[N-side-1][N-i-1];
            int left = matrix[N-i-1][side];
            matrix[i][N-side-1] = top;
            matrix[N-side-1][N-i-1] = right;
            matrix[N-i-1][side] = bottom;
            matrix[side][i] = left;
        }
    }
}