[C++]LeetCode: 109 Swap Nodes in Pairs (交换相邻节点位置)

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.

思路：

利用插入的办法，交换两个节点。维护一个pre节点，表示插入的前一个节点。维护一个cur节点，表示要插入的节点。同时构造一个虚拟头结点dummyhead，维护链表的头结点。

Attention:

1. 如果链表为空，或者只有一个节点，返回链表。

if(head == NULL || head->next == NULL) return head;

ListNode* dummyhead = new ListNode(0);
dummyhead->next = head;
ListNode* cur = head;
ListNode* pre = dummyhead;

while(cur != NULL && cur->next != NULL)

AC Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        ListNode* dummyhead = new ListNode(0);
        dummyhead->next = head;
        ListNode* cur = head;
        ListNode* pre = dummyhead;
        
        if(head == NULL || head->next == NULL) return head;
        while(cur != NULL && cur->next != NULL)
        {
            cur = cur->next;
            ListNode* tmp = cur->next;    //保存插入节点的后一个结点
            cur->next = pre->next;        //将cur节点插入
            pre->next = cur;              
            pre = cur->next;              //移动pre到下一个插入位置
            cur->next->next = tmp;        //连接交换后的节点和插入节点的后一个节点
            cur = tmp;                    //移动cur节点
        }
        return dummyhead->next;
    }
};

[C++]LeetCode: 109 Swap Nodes in Pairs (交换相邻节点位置)

原文：http://blog.csdn.net/cinderella_niu/article/details/42836767