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]forjinra
AI代码解释 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++;...
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){ ...
https://leetcode.com/problems/rotate-array/ class Solution { public: void rotate(vector<int>& nums, int k) { vector<int> nums_temp; k = k % nums.size(); for (vector<int>::size_type i = 0; i < nums.size(); i++) nums_temp.push_back(nums[(nums.size() - k + i) % nu...
1classSolution {2public:3voidpermute(vector<int>& nums,intstart,intend)4{5if(start >end)6return;78intsz = (end - start) /2+1;9for(inti =0; i < sz; i++)10{11inttmp = nums[start+i];12nums[start+i] = nums[end-i];13nums[end-i] =tmp;14}15}1617voidrotate(vector<int>&...
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...
力扣leetcode-cn.com/problems/rotate-array/ 题目描述 给定一个数组,将数组中的元素向右移动 k 个位置,其中 k 是非负数。 示例1: 输入: [1,2,3,4,5,6,7] 和 k = 3 输出: [5,6,7,1,2,3,4] 解释: 向右旋转 1 步: [7,1,2,3,4,5,6] 向右旋转 2 步: [6,7,1,2,3,4,5] ...
class Solution { public void rotate(int[] nums, int k) { if (nums == null || nums.length == 0 || k <= 0) return; int length = nums.length; int lastK = 0; if (k < length) lastK = k; else lastK = k % length; ...
class Solution { public void rotate(int[] nums, int k) { if (nums == null || nums.length == 0 || k <= 0) return; int length = nums.length; int lastK = 0; if (k < length) lastK = k; else lastK = k % length; ...
public class Solution { public void rotate(int[] nums, int k) { k = k % nums.length; reverse(nums, 0, nums.length-1); reverse(nums, 0, k-1); reverse(nums, k, nums.length-1); } private void reverse(int[] nums, int i, int j){ ...