leetCode 61.Rotate List (旋转链表) 解题思路和方法

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.

思路：题目很清晰，思路是先得到链表长度，再从头开始直到特定点，开始变换连接即可。

代码如下：

/**
 * 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(k == 0 || head == null)
            return head;
        int len = 0;
        ListNode first = head;//头结点
        ListNode last = null;//尾节点
        while(head != null){
            len++;//求长度
            last = head;//最后一个节点
            head = head.next;//循环条件
        }

        k = k%len;//如果k>len，取余数
        int n = 0;
        head = first;//标记到头结点
        while(head!= null){
            if(++n == (len - k))//判断是否到达位置
                break;
            head = head.next;
        }
        //以下为交换位置
        last.next = first;
        first = head.next;
        head.next = null;
        return first;
    }
}/**
 * 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(k == 0 || head == null)
            return head;
        int len = 0;
        ListNode first = head;//头结点
        ListNode last = null;//尾节点
        while(head != null){
            len++;//求长度
            last = head;//最后一个节点
            head = head.next;//循环条件
        }

        k = k%len;//如果k>len，取余数
        int n = 0;
        head = first;//标记到头结点
        while(head!= null){
            if(++n == (len - k))//判断是否到达位置
                break;
            head = head.next;
        }
        //以下为交换位置
        last.next = first;
        first = head.next;
        head.next = null;
        return first;
    }
}

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