给定一个数组(array),将数组向右旋转k步(k>=0)。 不需要输出,只改变原来的数组就可以。 Solution 1: 每次循环将数组的元素都向后移动一位,循环k次。 classSolution:defrotate(self, nums: List[int], k: int) ->None:foriinrange(k): previous= nums[-1]forjinrange(len(nums)): nums[j], previou...
1. Divide the array two parts: 1,2,3,4 and 5, 6 2. Reverse first part: 4,3,2,1,5,6 3. Reverse second part: 4,3,2,1,6,5 4. Reverse the whole array: 5,6,1,2,3,4 classSolution {publicvoidrotate(int[] nums,intk) {if(nums ==null){return; }if(k >=nums.length){ ...
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...
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. 将固定数组循环右移k步 注意:当k>numsSize的时候k = k % numsSize...leetcode - Rotate Array leetcode - Rotate Array Rotate an array of n elements to the right by k ...
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 k 个数字 Soluti...
189. 旋转数组(Rotate Array) 189. 旋转数组(Rotate Array) 题解 三次反转 复杂度分析 Python Java(待完成) 一些pythonic的解法 Java(待完成) 题解 哈哈,这题是我考研的数据结构编程题。最反映算法本质的做法是:三次反转。当然Python的语言特性解决这道题目也有很多办法。不过建议掌握三次反转。 三次反转 ...
Rotate Array Total Accepted: 12759 Total Submissions: 73112 My Submissions Question Solution Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. ...
class Solution: def rotate(self, nums, k): """ Do not return anything, modify nums in-place instead. """ n = len(nums) k %= n nums[:] = nums[::-1] nums[:k] = nums[:k][::-1] nums[k:] = nums[k:][::-1] LeetCode这个代码检验过程我有点搞不懂了,前两道题都return...
LeetCode Rotate Array 翻转数组 题意:给定一个数组,将该数组的后k位移动到前n-k位之前。(本题在编程珠玑中第二章有讲) 思路: 方法一:将后K位用vector容器装起来,再移动前n-k位到后面,再将容器内k位插到前面。 1 class Solution { 2 public:...
classSolution{publicvoidrotate(int[] nums,int k){intn= nums.length;int[] result =newint[nums.length];for(inti=; i < nums.length; i++) result[(i + k)% n]= nums[i];System.arraycopy(result,, nums,, n);}} 4.2> 翻转数组 classSolution{publicvoidrotate(int[] nums,int k){...