原題網(wǎng)址：https://www.hackerrank.com/challenges/matrix-rotation-algo

You are given a 2D matrix,?a, of dimension?MxN?and a positive integer?R. You have to rotate the matrix?R?times and print the resultant matrix. Rotation should be in anti-clockwise direction.

Rotation of a?4x5?matrix is represented by the following figure. Note that in one rotation, you have to shift elements by one step only (refer sample tests for more clarity).

It is guaranteed that the minimum of?M?and?N?will be even.

Input Format?
First line contains three space separated integers,?M,?N?and?R, where?M?is the number of rows,?N?is number of columns in matrix, and?R?is the number of times the matrix has to be rotated.?
Then?M?lines follow, where each line contains?N?space separated positive integers. These?M?lines represent the matrix.

Output Format?
Print the rotated matrix.

Constraints?
2 <=?M,?N?<= 300?
1 <=?R?<= 10⁹?
min(M, N) % 2 == 0?
1 <=?a_ij?<= 10⁸, where?i?∈?[1..M]?&?j?∈?[1..N]

Sample Input #00

4 4 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Sample Output #00

2 3 4 8 1 7 11 12 5 6 10 16 9 13 14 15

Sample Input #01

4 4 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Sample Output #01

3 4 8 12 2 11 10 16 1 7 6 15 5 9 13 14

Sample Input #02

5 4 7 1 2 3 4 7 8 9 10 13 14 15 16 19 20 21 22 25 26 27 28

Sample Output #02

28 27 26 25 22 9 15 19 16 8 21 13 10 14 20 7 4 3 2 1

Sample Input #03

2 2 3 1 1 1 1

Sample Output #03

1 1 1 1

Explanation?
Sample Case #00:?Here is an illustration of what happens when the matrix is rotated once.

1 2 3 4 2 3 4 85 6 7 8 1 7 11 129 10 11 12 -> 5 6 10 16 13 14 15 16 9 13 14 15

Sample Case #01:?Here is what happens when to the matrix after two rotations.

1 2 3 4 2 3 4 8 3 4 8 125 6 7 8 1 7 11 12 2 11 10 169 10 11 12 -> 5 6 10 16 -> 1 7 6 15 13 14 15 16 9 13 14 15 5 9 13 14

Sample Case #02:?Following are the intermediate states.

1 2 3 4 2 3 4 10 3 4 10 16 4 10 16 22 7 8 9 10 1 9 15 16 2 15 21 22 3 21 20 28 13 14 15 16 -> 7 8 21 22 -> 1 9 20 28 -> 2 15 14 27 -> 19 20 21 22 13 14 20 28 7 8 14 27 1 9 8 26 25 26 27 28 19 25 26 27 13 19 25 26 7 13 19 2510 16 22 28 16 22 28 27 22 28 27 26 28 27 26 254 20 14 27 10 14 8 26 16 8 9 25 22 9 15 193 21 8 26 -> 4 20 9 25 -> 10 14 15 19 -> 16 8 21 132 15 9 25 3 21 15 19 4 20 21 13 10 14 20 71 7 13 19 2 1 7 13 3 2 1 7 4 3 2 1

Sample Case #03:?As all elements are same, any rotation will reflect the same matrix.

方法：與一維數(shù)組平移相同，只需要將一維坐標(biāo)映射到二維。

import java.io.*; import java.util.*;public class Solution {private static int getY(int left, int top, int right, int bottom, int pos) {if (pos < bottom - top) return top + pos;if (pos < bottom - top + right - left) return bottom;if (pos < (bottom - top) * 2 + right - left) return top + ((bottom - top) * 2 + (right - left)) - pos;return top; }private static int getX(int left, int top, int right, int bottom, int pos) {if (pos < bottom - top) return left;if (pos < bottom - top + right - left) return left + pos - (bottom - top);if (pos < (bottom - top) * 2 + right - left) return right;return right - (pos - ((bottom - top) * 2 + (right - left)));}private static void reverse(int[][] matrix, int left, int top, int right, int bottom, int from, int to) {for(int i = from, j = to; i < j; i++, j--) {int iy = getY(left, top, right, bottom, i);int ix = getX(left, top, right, bottom, i);int jy = getY(left, top, right, bottom, j);int jx = getX(left, top, right, bottom, j);int t = matrix[iy][ix];matrix[iy][ix] = matrix[jy][jx];matrix[jy][jx] = t;}}private static void rotate(int[][] matrix, int left, int top, int right, int bottom, int rotate) {int length = (right - left) * 2 + (bottom - top) * 2;rotate %= length;if (rotate == 0) return;reverse(matrix, left, top, right, bottom, 0, length - 1 - rotate);reverse(matrix, left, top, right, bottom, length - rotate, length - 1);reverse(matrix, left, top, right, bottom, 0, length - 1); }public static void main(String[] args) {/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */Scanner scanner = new Scanner(System.in);int m = scanner.nextInt();int n = scanner.nextInt();int r = scanner.nextInt();int[][] matrix = new int[m][n];for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {matrix[i][j] = scanner.nextInt();}}int left = 0, top = 0, right = n - 1, bottom = m - 1;while (left < right && top < bottom) {rotate(matrix, left, top, right, bottom, r);left++;top++;right--;bottom--;}for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {if (j > 0) System.out.print(" ");System.out.print(matrix[i][j]);}System.out.println();}} }

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