Leetcode:Swap Nodes in Pairs 单链表相邻两节点逆置

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.

/**
 * 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) {
        if(head == NULL || head->next == NULL)
            return head;
        ListNode dummy(0);
        ListNode *tail = &dummy;
        ListNode *pre = head;
        ListNode *cur = NULL;
        ListNode *nxt = NULL;
        
        while(pre != NULL)
        {
            cur = pre->next;
            if(cur != NULL)
            {
                nxt = cur->next;
                cur->next = pre;
                pre->next = NULL;
                tail->next = cur;
                tail = pre;
                pre = nxt;
            }
            else
            {
                tail->next = pre;
                return dummy.next;
            }
            
        }
        
        return dummy.next;
    }
};

Leetcode:Swap Nodes in Pairs 单链表相邻两节点逆置,布布扣,bubuko.com

Leetcode:Swap Nodes in Pairs 单链表相邻两节点逆置

原文：http://blog.csdn.net/u012118523/article/details/37721675