24. Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list,
only nodes itself can be changed.

关于链表的题, 画图看看需要那几个节点, 一般都是接头节点(用作标记) 和遍历节点, 遍历的时候要判空, head.next !=  null 前一定要判head != null, 

将尾节点置为空!

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy =  new ListNode(0);
        ListNode pre = dummy;
        while (head != null && head.next != null) {
            ListNode temp = head.next.next;
            pre.next = head.next;
            pre = pre.next;
            pre.next = head;
            pre = pre.next;
            head = temp;
        }
        if (head != null) {
            pre.next = head;
            pre.next.next = null;
        } else {
             pre.next = null;
        }
       
        return dummy.next;
        
    }
}

  

原文地址:https://www.cnblogs.com/apanda009/p/7396782.html