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]
]
我剛開始想到的是比較暴力的算法慕爬，即把所有數(shù)字都用list保存起來，然后再讓他們回到各自的位置：
public void rotate(int[][] matrix) {
int length = matrix.length;
List<Integer> list = new ArrayList<>();
for (int i = 0; i < length; i++) {
for (int j = 0; j < length; j++) {
list.add(matrix[i][j]);
}
}
for (int i = 0; i < length; i++) {
for (int j = length - 1; j >= 0; j--) {
matrix[j][i] = list.remove(list.size() - 1);
}
}
}
還有一種思路是狮鸭，四個位置輪流交換：
思路圖
思路就是灭衷，四個位置相互交換一下就可以放前。
public class Solution {
public void rotate(int[][] matrix) {
for(int i=0, temp=0, n=matrix.length-1; i<=n/2; i++) {
for(int j=i; j<n-i; j++) {
temp = matrix[j][n-i];
matrix[j][n-i] = matrix[i][j];
matrix[i][j] = matrix[n-j][i];
matrix[n-j][i] = matrix[n-i][n-j];
matrix[n-i][n-j] = temp;
}
}
}
}
