[LeetCode] Swap Nodes in Pairs

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题目：

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.

题解：

算法很直观，直接一次遍历就行了，要多注意下指针的移动，为了方便处理，定义了一个头指针

代码如下：

/** * 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 *h= (ListNode *)malloc(sizeof(ListNode));h->next = head;if (head == NULL || head->next == NULL){return head;}ListNode *pre = h;ListNode *ptr = pre->next;ListNode *tail = ptr->next;while ( true ){pre->next = tail;ptr->next = tail->next;tail->next = ptr;pre = ptr;if (ptr->next)ptr = ptr->next;elsebreak;if (ptr->next)tail = ptr->next;elsebreak;}return h->next;}};

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