forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_48.java
81 lines (72 loc) · 2.67 KB
/
_48.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
package com.fishercoder.solutions;
/**
* 48. Rotate Image
*
* You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Follow up:
Could you do this in-place?
*/
public class _48 {
/**Note: this is an n*n matrix, in other words, it's a square, this makes it easier as well.*/
public static class Solution1 {
public void rotate(int[][] matrix) {
/**First swap the elements on the diagonal, then reverse each row:
* 1, 2, 3 1, 4, 7 7, 4, 1
* 4, 5, 6 becomes 2, 5, 8 becomes 8, 5, 2
* 7, 8, 9 3, 6, 9 9, 6, 3
This is done in O(1) space!
**/
int m = matrix.length;
for (int i = 0; i < m; i++) {
for (int j = i; j < m; j++) {
/**ATTN: j starts from i, so that the diagonal changes with itself, results in no change.*/
int tmp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = tmp;
}
}
/**then reverse*/
for (int i = 0; i < m; i++) {
int left = 0;
int right = m - 1;
while (left < right) {
int tmp = matrix[i][left];
matrix[i][left] = matrix[i][right];
matrix[i][right] = tmp;
left++;
right--;
}
}
}
}
public static class Solution2 {
/**First swap the rows bottom up, then swap the element on the diagonal:
* 1, 2, 3 7, 8, 9 7, 4, 1
* 4, 5, 6 becomes 4, 5, 6 becomes 8, 5, 2
* 7, 8, 9 1, 2, 3 9, 6, 3
*
* This is using O(n) of extra space
*/
public void rotate(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
int top = 0;
int bottom = n - 1;
while (top < bottom) {
int[] tmp = matrix[top];
matrix[top] = matrix[bottom];
matrix[bottom] = tmp;
top++;
bottom--;
}
for (int i = 0; i < m; i++) {
for (int j = i + 1; j < n; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = tmp;
}
}
}
}
}