Swap Nodes in Pairs

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

You may not modify the values in the list's nodes, only nodes itself may be changed.

 

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

/* 两两交换链表节点
 * 1 2 3 4 -> 2 1 4 3
 * 方法： 设立头结点避免特判 每次循环完成一组交换 并且维持一个next
 * */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(!head)   return NULL;
        // 问题 第一次进行交换时已经对head进行重新赋值了()
        ListNode *pre=new ListNode(0), *p=head, *tmp, *newhead=head;
        pre->next = head; // pre为当前要交换节点的前置节点
        if(head->next)  newhead = head->next;
        while(p && p->next){// 节点要有两个时才进行交换
            tmp = p->next->next;
            pre->next = p->next;
            pre->next->next = p;
            p->next = tmp;
            pre = p;
            p = tmp;
        }
        return newhead;

    }
};