24. Swap Nodes in Pairs

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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list的操作，下面是我的错误答案，input [1,2]，output[1]，excepted[2,1]。

public class Solution {    public ListNode swapPairs(ListNode head) {        if (head == null) {            return null;        }        ListNode headlist = new ListNode(0);        ListNode templist = new ListNode(0);        headlist.next = head;        while (head.next != null) {            templist.next = head.next.next;            head.next.next = head;            head.next = templist.next;            head = head.next;            if (head == null) {                break;            }            if (head.next == null) {                break;            }        }         return headlist.next;    }}

下面是正确的做法：

public ListNode swapPairs(ListNode head) {    ListNode dummy = new ListNode(0);    dummy.next = head;    ListNode current = dummy;    while (current.next != null && current.next.next != null) {        ListNode first = current.next;        ListNode second = current.next.next;        first.next = second.next;        current.next = second;        current.next.next = first;        current = current.next.next;    }    return dummy.next;}

可以看到我是直接对形参head操作，而正确做法是新建current，first，second，对所有要用到的listnode都新建，暂时还不知道为什么这么做。

知道了，这里不可以改head，因为代码中把headlist.next设成head，就不要动head了，否则结果会错。不过上面的代码明显有个瑕疵：

public class Solution {    public ListNode swapPairs(ListNode head) {        ListNode headlist = new ListNode(0);        ListNode curlist = new ListNode(0);        headlist.next = head;        curlist = headlist;        while (curlist.next != null && curlist.next.next != null) {            ListNode first = curlist.next;            ListNode second = curlist.next.next;            first.next = second.next;            second.next = first; //这里应该改变第二个元素的next啊，不知道系统为什么认为不加也对            curlist.next = second;            curlist.next.next = first;            curlist = curlist.next.next;        }         return headlist.next;    }}

这道题的递归解法也很奇妙，先记录下原本的head.next，然后head.next 其实应该等于swamPairs(head.next.next)，然后把原本的head.next.next设成head，代码如下：

public class Solution {    public ListNode swapPairs(ListNode head) {        if ((head == null)||(head.next == null))            return head;        ListNode n = head.next;        head.next = swapPairs(head.next.next);        n.next = head;        return n;    }}

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