Leetcode-Rotate Array（189 ，61）

Rotate Array（189 ，61）

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].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Related problem: Reverse Words in a String II

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

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三步：全部反转；反转序列1 ；反转序列2

public class Solution {
    public void rotate(int[] nums, int 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;
            start++;
            end--;
        }
    }

}

Rotate List（61）

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

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Since n may be a large number compared to the length of list. So we need to know the length of linked list.After that, move the list after the (l-n%l )th node to the front to finish the rotation.

Ex: {1,2,3} k=2 Move the list after the 1st node to the front

Ex: {1,2,3} k=5, In this case Move the list after (3-5%3=1)st node to the front.

So the code has three parts.

1. Get the length
2. Move to the (l-n%l)th node

3)Do the rotation

public ListNode rotateRight(ListNode head, int n) {
    if (head==null||head.next==null) return head;
    ListNode dummy=new ListNode(0);
    dummy.next=head;
    ListNode fast=dummy,slow=dummy;

    int i;
    for (i=0;fast.next!=null;i++)//Get the total length 
        fast=fast.next;
    
    for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
        slow=slow.next;
    
    fast.next=dummy.next; //Do the rotation
    dummy.next=slow.next;
    slow.next=null;
    
    return dummy.next;
}