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  • [LeetCode 48] Rotate Image

    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

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  • 原文地址:https://www.cnblogs.com/lz87/p/7056900.html
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