24. Swap Nodes in Pairs

Problem:

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.

5/1/2018 update:
First.next is like next in regular reverse. 

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Analysis:

Simple reverse linked list, but need to think about how to handle head. dummy不是随便一初始化就行了，需要考虑dummy的安置问题，不留死角。

Solution:

cur.next 把head给转移到了second, 然后cur就完成了任务，继续下一次reverse。

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null)
            return null;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        while (prev.next != null && prev.next.next != null) {
            ListNode first = prev.next, second = prev.next.next;
            first.next = second.next;
            second.next = first;
            prev.next = second;
            prev = first;
        }
        return dummy.next;
    }
}