[leetcode]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.

这道题我感觉最重要的点是在链表头再加一个临时结点，这样就不用去判断那些特殊情况了。

Java代码

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode swapPairs(ListNode head) {        ListNode cur = head;        ListNode tmp = null,preHead=new ListNode(0);        preHead.next=head;        ListNode pre=preHead;        ListNode node1=null,node2=null;        while((node1=cur)!=null&&(node2=cur.next)!=null){            tmp = node2.next;            node2.next=node1;            node1.next=tmp;            pre.next=node2;            pre=node1;            cur=tmp;                    }        return preHead.next;    }}

go代码

/** * Definition for singly-linked list. * type ListNode struct { *     Val int *     Next *ListNode * } */func swapPairs(head *ListNode) *ListNode {    preHead:=&ListNode{        Val:0,        Next:head,    }    var node1,node2,cur,tmp,pre *ListNode    cur=head    pre=preHead    for (cur!=nil)&&(cur.Next!=nil){        node1=cur        node2=cur.Next        tmp=node2.Next        node2.Next=node1        node1.Next=tmp        pre.Next=node2        pre=node1        cur=tmp    }    return preHead.Next}