【LeetCode】C# 48、Rotate Image

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最新推荐文章于 2024-10-22 15:01:03 发布

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本文介绍了一种在不使用额外内存的情况下，将一个n×n的二维矩阵顺时针旋转90度的方法。提供了两种解决方案，一种是通过创建新的二维数组来实现，另一种则是通过在原地修改输入矩阵来完成旋转。

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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]
]

将给定矩阵顺时针翻转90°。
思路1：直接旋转，要实例化一个新的二维数组，能AC但不合题意。。

public class Solution {
    public void Rotate(int[,] matrix) {
        int n = matrix.GetLength(0);
        int[,] temp = new int[n, n];
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                temp[i, j] = matrix[n - j -1, i];
            }
        }
        for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    matrix[i, j] = temp[i, j];
                }
            }
    }
}

思路2：借助一个temp整数，把图一圈一圈替换，实现inplace

int i,j,temp;  
        int n=matrix.GetLength(0);  
        for(i = 0;i < n/2;++i) {  
            for (j = i;j < n-1-i;++j) {  
                temp = matrix[j][n-i-1];  
                matrix[j][n-i-1] = matrix[i][j];  
                matrix[i][j] = matrix[n-j-1][i];  
                matrix[n-j-1][i] = matrix[n-i-1][n-j-1];  
                matrix[n-i-1][n-j-1] = temp;  
            }  
        }