[LeetCode] 48. Rotate Image
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.
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] ]
Thought process:
Iterate the matrix from the outer-most layer to the inner most layer. Rotate the layers one by one.
Solution (C++):
1 2 3 4 5 6 7 8 9 10 11 12 | class Solution { public: void rotate(vector<vector<int>>& matrix) { reverse(matrix.begin(), matrix.end()); for (int i = 0; i < matrix.size(); i++) { for (int j = i; j < matrix.size(); j++) { swap(matrix[i][j], matrix[j][i]); } } } }; |
Solution (Java):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | class Solution { public void rotate(int[][] matrix) { for (int i = 0; i < matrix.length / 2; i++) { for (int j = i; j < matrix.length - i - 1; j++) { rotate(matrix, i, j); } } } private void rotate(int[][] matrix, int i, int j) { int n = matrix.length; int temp = matrix[i][j]; matrix[i][j] = matrix[n - 1 - j][i]; matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; matrix[j][n - 1 - i] = temp; } } |
Time complexity: O(n^2), where n is the length of the matrix.
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