leetcode: Swap Nodes in Pairs

http://oj.leetcode.com/problems/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.

思路：

每次读两个节点插入新链表，因为是指针操作，不占用额外空间。最后需要把tail指针的next设为NULL。例如链表为p1->p2->p3->p4的情况，如果最后不修改p3->next，那么就会出现一个环：p2->p1->p4->p3->p4->...

class Solution {
public:
    void insertTwoNodes(ListNode *p1, ListNode *p2, ListNode *&head, ListNode *&tail) {
        if (NULL == head) {
            head = tail = p2;
        }
        else {
            if (p2 != NULL) {
                tail->next = p2;
                tail = p2;
            }
        }
        
        if (NULL == head) {
            head = tail = p1;
        }
        else {
            tail->next = p1;
            tail = p1;
        }
    }
    
    ListNode *swapPairs(ListNode *head) {
        ListNode *swapped_head = NULL, *swapped_tail = NULL;
        
        while (head != NULL) {
            ListNode *p1 = NULL, *p2 = NULL;
            
            p1 = head;
            head = head->next;
            
            if (head != NULL) {
                p2 = head;
                head = head->next;
            }
            
            insertTwoNodes(p1, p2, swapped_head, swapped_tail);
        }
        
        if (NULL != swapped_tail) {
            swapped_tail->next = NULL;
        }
        
        return swapped_head;
    }
};