[LeetCode 48] Rotate Image

2020-09-24 05:32:06 阅读：195 来源： 互联网

标签：Rotate last matrix int Image top length rotate LeetCode

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

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:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Example 3:

Input: matrix = [[1]]
Output: [[1]]

Example 4:

Input: matrix = [[1,2],[3,4]]
Output: [[3,1],[4,2]]

Constraints:

matrix.length == n
matrix[i].length == n
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000

Solution 1. O(n^2) runtime, O(n^2) space

 1 public class Solution {
 2     public void rotate(int[][] matrix) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return;
 5         }
 6         int n = matrix.length;
 7         int[][] rotate = new int[n][n];
 8         for(int i = 0; i < n; i++){
 9             for(int j = 0; j < n; j++){
10                 rotate[i][j] = matrix[n - 1 - j][i];
11             }
12         }
13         for(int i = 0; i < n; i++){
14             for(int j = 0; j < n; j++){
15                 matrix[i][j] = rotate[i][j];
16             }
17         }
18         //matrix = rotate;
19     }
20 }

One pitfall here is that line 18 does not change the original matrix that needs to be rotated. Java is pass-by-copy, so a copy of matrix, which is

a reference to the actual memory location of the 2D matrix, is passed to method rotate. You should use this copy of reference to change the

actual matrix entries. Changing this copy of reference to a new location does not touch the original matrix at all.

Solution 2. O(n^2) runtime, O(1) space

The runtime in solution 1 is already BCR asymptotically. But we can still optimize on extra memory usage.

Perform a circular rotation on each layer,(left -> top, bottom -> left, right -> bottom, top -> bottom. ),

starting from the outermost layer and working our way inwards.

How do we perform this 4-way edge swap? One option is to copy the top edge to an array, then move

the left to the top, the bottom to the left, and so on. This requires O(n) memory.

A better way to do this is to implement the swap index by index, so we only needs O(1) extra memory.

 1 public class Solution {
 2     public void rotate(int[][] matrix) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return;
 5         }
 6         int n = matrix.length;
 7         for(int layer = 0; layer < n / 2; layer++){
 8             int first = layer;
 9             int last = n - 1 - layer;
10             for(int i = first; i < last; i++){
11                 int offset = i - first;
12                 //save top
13                 int top = matrix[first][i];
14                 //left -> top
15                 matrix[first][i] = matrix[last - offset][first];
16                 //bottom -> left
17                 matrix[last - offset][first] = matrix[last][last - offset];
18                 //right -> bottom
19                 matrix[last][last - offset] = matrix[i][last];
20                 //top -> right
21                 matrix[i][last] = top;
22             }
23         }
24     }
25 }

Related Problems

Rotate String

Matrix Zigzag Traversal

Set Matrix Zeroes

标签：Rotate,last,matrix,int,Image,top,length,rotate,LeetCode
来源： https://www.cnblogs.com/lz87/p/7056900.html

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[LeetCode 48] Rotate Image