代码(Python3) class Solution: def rotate(self, nums: List[int], k: int) -> None: """ Do not return anything, modify nums in-place instead. """ n: int = len(nums) # 计算最后有多少数字会被移动到数组开始 k %= n # 翻转整个数组 Solution.reverse(nums, 0, n - 1) # 翻转前 ...
Runtime: 48 ms, faster than 94.32% of Python3 online submissions forRotate Array. Memory Usage: 13.7 MB, less than 5.23% of Python3 online submissions for Rotate Array. 将代码优化: class Solution: def rotate(self, nums: List[int], k: int) -> None: """ Do not return anything, modi...
Special thanks to@Freezenfor adding this problem and creating all test cases. AC代码:(Python) 1classSolution:2#@param nums, a list of integer3#@param k, num of steps4#@return nothing, please modify the nums list in-place.5defrotate(self, nums, k):6n =len(nums)7k = k %n8nums[:...
代码运行次数:0 publicclassSolution{publicvoidrotate(int[]nums,int k){k%=nums.length;int[]rotateNums=newint[nums.length];int index=0;for(int i=nums.length-k;i<nums.length;i++){rotateNums[index]=nums[i];index++;}for(int i=0;i<nums.length-k;i++){rotateNums[index]=nums[i];index...
代码二 (python) 由于是旋转数组,所以二分查找更新边界的时候多了限制。如果中间的数小于最右边的数,则右半段是有序的,若中间数大于最右边数,则左半段是有序的,我们只要在有序的半段里用首尾两个数来判断目标值是否在这一区域内。 class Solution: ...
Can you solve this real interview question? Search in Rotated Sorted Array - There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1
如何使用Python解决Leetcode的Rotate Image问题? 题目: 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? 思路分析: 最笨的方法,重新开辟一个矩阵空间,做旋转。(题目要求最好能就地旋转) 更好的方法:先将矩...
[Array]Rotate Array Problem link: https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/646/leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/646/ Solutions: Given an array, rotate the array to the right byksteps, wherekis non-negative....
在python中很容易使用切片操作达到循环移位的做法。只需要循环A长度次,看看每个结果即是所有的可能的移位结果。 class Solution(object): def rotateString(self, A, B): """ :type A: str :type B: str :rtype: bool """ if A == B == "": ...
AI代码解释 classSolution{public:voidrotate(vector<vector<int>>&matrix){if(matrix.size()<=0)return;int a=0,b=matrix.size()-1;while(a<b){for(int i=0;i<b-a;++i){swap(matrix[a][a+i],matrix[a+i][b]);swap(matrix[a][a+i],matrix[b][b-i]);swap(matrix[a][a+i],matrix[...