2.2.6 Rotate List

Link: https://oj.leetcode.com/problems/rotate-list/

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.

我的思路：

1 找出数组长度。2 找出断开位置。3 把old tail连上old head。但是做错了。

我的第一遍代码：

public class Solution {    public ListNode rotateRight(ListNode head, int n) {        if(head == null || head.next == null || n == 0) return head;        //first find out the length of the list        ListNode cur = head;        int len = 0;        while(cur != null){            cur = cur.next;            len ++;//len = 5        }        //move cur to the new head        cur = head;        for(int i = 1; i < len - n; i++){            cur = cur.next;        }        ListNode newHead = cur.next;        cur.next = null;        ListNode tail = newHead;        while(tail.next != null){            tail = tail.next;        }        tail.next = head;        return newHead;    }}

Input:{1,2}, 2Output:{2,1}Expected:{1,2}

正确的思路：上述第2， 3步交换顺序。1 找出数组长度。2 把old tail连上old head。3 断开。

注意：n 可能大于链表长度，所以要用len - n % len （而不是len - n)

Time: O(n), Space: O(1)

public class Solution {    public ListNode rotateRight(ListNode head, int n) {        if(head == null || head.next == null || n == 0) return head;        //first find out the length of the list        ListNode cur = head;        int len = 1;        while(cur.next != null){            len ++;            cur = cur.next;        }        //connect tail to head to make a circle        cur.next = head;        for(int i = 0; i < len - n%len; i++){            cur = cur.next;        }        ListNode newHead = cur.next;        cur.next = null;        return newHead;    }}