Rotate List —— LeetCode

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.

 

题目大意:给一个链表和一个非负整数k,把链表向右循环移位k位。

解题思路:踩了个坑,k有可能比链表的长度还长,比如长度为3的链表,k=2000000000,所以开始需要遍历一下链表算出长度,k%=len,然后就是两个pointer一个先走k步,另一个再走,最后把末尾的next节点设为head,新head就是慢指针指向的节点。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null||k == 0){
            return head;
        }
        int fast = k;
        ListNode fastNode = head;
        ListNode slowNode = head;
        ListNode pre = slowNode;
        int len = 0;
        while(fastNode!=null){
            len++;
            fastNode=fastNode.next;
        }
        fast %= len;
        fastNode = head;
        if(fast == 0){
            return head;
        }
        while(fastNode!=null&&fast-->0){
            fastNode = fastNode.next;
        }
        while(fastNode!=null){
            fastNode = fastNode.next;
            pre = slowNode;
            slowNode = slowNode.next;
        }
        ListNode newHead = new ListNode(0);
        newHead.next = slowNode;
        pre.next = null;
        while(slowNode.next!=null){
            slowNode = slowNode.next;
        }
        slowNode.next = head;
        return newHead.next;
    }
}

 

posted @ 2015-05-20 13:49  丶Blank  阅读(127)  评论(0编辑  收藏  举报