
问题
给定一个 n × n 的二维矩阵表示一个图像。
将图像顺时针旋转 90 度。
说明:
你必须在原地旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。
给定 matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],原地旋转输入矩阵,使其变为:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
给定 matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],原地旋转输入矩阵,使其变为:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
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.
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]
]
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]
]
示例
public class Program { public static void Main(string[] args) { var matrix = new int[,] { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; Rotate(matrix); ShowArray(matrix); matrix = new int[,] { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; Rotate2(matrix); ShowArray(matrix); Console.ReadKey(); } private static void ShowArray(int[,] matrix) { var length = matrix.GetLength(0); for(var i = 0; i < length; i++) { for(var j = 0; j < length; j++) { Console.Write($"{ matrix[i, j]} "); } Console.WriteLine(); } Console.WriteLine(); } public static void Rotate(int[,] matrix) { var length = matrix.GetLength(0); for(var i = 0; i < length; i++) { for(var j = i; j < length; j++) { var swap = matrix[i, j]; matrix[i, j] = matrix[j, i]; matrix[j, i] = swap; } } for(var i = 0; i < length; i++) { for(var j = 0; j < length / 2; j++) { var swap = matrix[i, j]; matrix[i, j] = matrix[i, length - j - 1]; matrix[i, length - j - 1] = swap; } } } public static void Rotate2(int[,] matrix) { var length = matrix.GetLength(0); for(var i = 0; i < length / 2; i++) { for(var j = i + 1; j < length - i; j++) { var swap = matrix[i, j]; matrix[i, j] = matrix[length - 1 - j, i]; matrix[length - 1 - j, i] = matrix[length - 1 - i, length - 1 - j]; matrix[length - 1 - i, length - 1 - j] = matrix[j, length - 1 - i]; matrix[j, length - 1 - i] = swap; } } } }
以上给出2种算法实现,以下是这个案例的输出结果:
7 4 1 8 5 2 9 6 3 13 9 5 1 14 10 6 2 15 11 7 3 16 12 8 4
分析:
显而易见,Rotate 和 Rotate2 的时间复杂度均为: 。
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