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.

这道题考察的是两个节点的交换，这里是指的是指针的交换，而不是值的交换，注意指针即可。

需要注意的地方是：设置head头结点，原因是要对第一队的交换做特殊处理，所以才添加一个head头节点。

代码如下：

/* class ListNode {      int val;      ListNode next;      ListNode(int x) { val = x; }  }*//* * 设置head结点，两两交换 *  * 设置head的原因是：考虑到第一对结点的交换， * 所以设置head结点，一遍统一处理 *  * */public class Solution {     public ListNode swapPairs(ListNode head)      {         if(head==null || head.next==null)             return head;         ListNode headListNode=new ListNode(-1);         headListNode.next=head;         ListNode fa=headListNode,now=fa.next,nextNow=now.next;         while(now!=null && nextNow!=null)         {             now.next=nextNow.next;             nextNow.next=now;             fa.next=nextNow;             fa=now;             now=fa.next;             nextNow=now!=null? now.next : null;         }            return headListNode.next;     }}

下面是C++代码，就是指针交换即可，

代码如下：

#include <iostream>using namespace std;/*struct ListNode{int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};*/class Solution{public:    ListNode* swapPairs(ListNode* head)     {        if (head == NULL || head->next == NULL)            return head;        ListNode* newHead = new ListNode(-1);        newHead->next = head;        ListNode* fa = newHead;        ListNode* now = fa->next;        ListNode* next = now->next;        while (now != NULL && next != NULL)        {            ListNode* tmp = next->next;            next->next = now;            fa->next = next;            now->next = tmp;            fa = now;            now = fa->next;            next = now == NULL ? NULL : now->next;        }        return newHead->next;    }};