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; } };