matrix[i], matrix[n - 1 - i] = matrix[n - 1 - i], matrix[i] for i in range(1,n): #Begin with '1' instead of 0 can avoide action on main-diagonal for j in range(i): #沿着主对角线翻转; matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] return matrix""...
You are given annxn2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Note: You have to rotate the imagein-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation. 翻译:你有一个n*n 的2D矩阵表...
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. 这道题就是输一个n*...
Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? 思路: 先将矩阵转置,再将矩阵按中轴线对称交换每一列。 算法: public void rotate(int[][] matrix) { for (int i = 0; i < matrix.length; i++) { //转置 for (int j = i; j < matrix[0].length;...
matrix[i].length == n 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000 Rotate Image Medium 4990343Add to ListShare You are given annxn2Dmatrixrepresenting an image, rotate the image by 90 degrees (clockwise). You have to rotate the imagein-place, which means you have to modify the...
You are given annxn2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Note: You have to rotate the imagein-place, which means you have to modify the input 2D matrix directly.DO NOTallocate another 2D matrix and do the rotation. ...
LeetCode:Rotate Image You are given annxn2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? 不使用额外的空间顺时针旋转方阵90度 例如 旋转后变为 算法1 先将矩阵转置,然后把转置后的矩阵每一行翻转...
Leetcode: Rotate Image Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place?...C++参考代码: class Solution { public: void rotate(vector > &matrix) { if (matrix.empty 66420 Rotate List 思路 先统计链表的长度n,如果k>n,就取k=k%n,如果k==0,就不用做变化...
/* * clockwise rotate 顺时针旋转 * first reverse up to down, then swap the symmetry * 1 2 3 7 8 9 7 4 1 * 4 5 6 => 4 5 6 => 8 5 2 * 7 8 9 1 2 3 9 6 3 */voidrotate(vector<vector<int>>&matrix){reverse(matrix.begin(),matrix.end());for(inti=0;i<matrix.size...
Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? In-place Solution By using the relation “matrix[i][j] = matrix[n-1-j][i]”, we can loop through the matrix. publicvoidrotate(int[][]matrix){intn=matrix.length;for(inti=0;i<n/2;i++){for(...