LeetCode Swap Nodes in Pairs交换链表中的两个节点

Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given1->2->3->4, you should return the list as2->1->4->3.

Your algorithm should use only constant space. You maynotmodify the values in the list, only nodes itself can be changed.

入门题，注意：

1 保存next节点

2 需要使用三个节点指针

3 保存好链表头指针

/**
 * 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 || !head->next) return head;
		ListNode *pre = head->next;
		ListNode *cur = pre->next;
		pre->next = head;
		head->next = cur;
		head = pre;
		pre = pre->next;
		while (pre)
		{
			ListNode *post;
			cur = pre->next;
			if (cur && cur->next) post = cur->next;
			else break;
			pre->next = post;
			pre = post->next;
			post->next = cur;
			cur->next = pre;
			pre = cur;
		}
		return head;
	}
};

2014-1-25 update

	ListNode *swapPairs(ListNode *h)
	{
		ListNode dummy(0);
		dummy.next = h;
		ListNode *pre = &dummy;
		while (h && h->next)
		{
			ListNode *t = h->next->next;
			pre->next = h->next;
			h->next->next = h;
			h->next = t;
			pre = h;
			h = t;
		}
		return dummy.next;
	}