189. 轮转数组
给定一个整数数组 nums,将数组中的元素向右轮转 k 个位置,其中 k 是非负数。
示例 1:
输入: nums = [1,2,3,4,5,6,7], k = 3 输出: [5,6,7,1,2,3,4]解释: 向右轮转 1 步: [7,1,2,3,4,5,6] 向右轮转 2 步: [6,7,1,2,3,4,5] 向右轮转 3 步: [5,6,7,1,2,3,4]
示例 2:
输入:nums = [-1,-100,3,99], k = 2 输出:[3,99,-1,-100] 解释: 向右轮转 1 步: [99,-1,-100,3] 向右轮转 2 步: [3,99,-1,-100]
提示:
1 <= nums.length <= 105-231 <= nums[i] <= 231 - 10 <= k <= 105
方法1:

public static void rotate(int[] nums, int k) {int length = nums.length;int[] newArr = new int[length];k = k > length? k % length : k;for (int i = 0; i < nums.length; i++) {newArr[i] = nums[(length + i - k) % length];}System.arraycopy(newArr, 0, nums, 0, length);}
方法2:
public void rotate(int[] nums, int k) {//翻转数组k = k % nums.length;reverse(nums,0,nums.length-1);reverse(nums,0,k-1);reverse(nums,k,nums.length-1);}public void reverse(int[] nums, int start,int end){while(start<end){int temp = nums[start];nums[start++] = nums[end];nums[end--] = temp;}}