小白笔记-------------------------------------------------（leetcode:24. Swap Nodes in Pairs）

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

Example:

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

Note:

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

这道题，主要是两个指针的操作，一个是头部指针，即head不断地往后移动，每次移动两个，一个则是每个pair的上一个指针，即prev，都要跟着移动，主要就是这两个指针的变化，如果检测到下一个pair不到两个节点，则停止操作

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode prev = new ListNode(0);
        prev.next = head;
        ListNode preMark = prev;
        while(head != null && head.next != null){
           ListNode sec = head.next;
           head.next = sec.next;
           sec.next = head;
           prev.next = sec;
           prev = head;
           head = head.next;
        }
        return preMark.next;
    }
}