matrix[i].reverse()#然后将矩阵的每一行翻转returnmatrix 解题思路2:首先沿着副对角线翻转一次,然后沿着水平中线翻转一次,就可以得到所要求的矩阵了。时间复杂度O(n^2),空间复杂度O(1) classSolution:#@param matrix, a list of lists of integers#@return a list of lists of integersdefrotate(self, matrix...
You are given annxn2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? 分析: 二维数组a[n][n]顺时针旋转90度,要解决这个问题,无疑,第一件事儿就是找规律。 当n=1时,不用动了。 当n=2时, 旋转之后变为 有: a[0][0] =...
classSolution:defrotate(self,matrix:List[List[int]])->None:""" Do notreturnanything,modify matrixin-place instead.""" n=len(matrix)# 注意一下范围和下标即可foriinrange(n//2):forjinrange(i,n-1-i):matrix[i][j],matrix[j][n-1-i],matrix[n-1-i][n-1-j],matrix[n-1-j][i]=\...
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? 本题难度Medium。 【题意】 对一个正方形矩阵进行顺时针旋转90度。 【复杂度】 时间O(N) 空间 O(1) 【思路】 第一步:镜像对换 如: 1 2 3 4 5...
【Leetcode】Rotate Image 题目: n x 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...
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 input 2D matrix directly.DO NOTallocate another 2D matrix and do the rotation. ...
classSolution{public:voidrotate(vector<vector<int>>&matrix){int n=matrix.size();for(int i=0;i<(n+1)/2;i++)// 行的遍历范围{for(int j=0;j<n/2;j++)// 列的遍历范围{// 由于是旋转赋值,所以temp记录的是最后一个位置上的元素//然后逆时针进行覆盖,见上面的图。int temp=matrix[n-1-...
Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? 分析 给定一个N*N的二维数组,求其顺时针旋转90度后的数组。 只要简单分析一下数字之间的对应关系即可。C代码如下已通过。 voidrotate(int**matrix,intmatrixRowSize,intmatrixColSize){int**ans=(int**)malloc(sizeof...
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...
原题链接:https://oj.leetcode.com/problems/rotate-list/ 得到链表长度。链表指向尾节点,将链表首尾相连,向右k%len。得到结果链表的尾节点。 publicclassRotateList{ publicstaticvoidmain(String[]args){ ListNodehead=newListNode(1); ListNodehead1=newListNode(2); ...