LeetCode 189. Rotating array (java implementation)

189. Rotate array

Given an array, rotate the elements in the array k positions to the right, where k is a non-negative number.

Example 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]

Example 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]

hint:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Answer ideas:

• Because there is no return value here, it is inconvenient to directly operate on the original array, so first make a copy of the original array.
• Shift all elements by k positions
• Just put the original element of arr[i] at the position of arr[(i+k)%arr.length]

class Solution {
    
    
    public void rotate(int[] nums, int k) {
    
    
    int []arr=new int[nums.length];//创建同等大小的新数组
    for(int i=0;i<nums.length;i++){
    
    
          arr[i]=nums[i];         //将原数组复制到新数组中，因为这里没有返回值，直接在原数组中操作不方便
         }
    int n=nums.length;          //记录数组长度
    for(int i=0;i<arr.length;i++){
    
    
    nums[(i+k)%n]=arr[i];//向右移动三个位置，第一个变成了第四个，以此类推，位移后的位置为当前位置+k,再对 数组长度取余
        }
    }
}