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.

这个问题我们先建一个头结点，这样可以统一操作，每一步就那么几个操作，见图。



这样就不用特别处理第一结点，好了看明白这个图说的步骤就代码。

class Solution {
public:
ListNode *swapPairs(ListNode *head) {
if(head == NULL || head->next == NULL)
return head;
ListNode *pre, *tail;
ListNode *new_head = new ListNode(-1);
new_head->next = head;
pre = new_head;
while(pre->next && pre->next->next)
{
tail = pre->next->next;
pre->next->next = tail->next;
tail->next = pre->next;
pre->next = tail;
pre = tail->next;
}
pre = new_head;
new_head = new_head->next;
delete pre;
return new_head;
}
};

leetcode里有好多链表题，这里是创建链表的函数，想自己测试的在这里下：创建链表函数