Swap Nodes in Pairs

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.

首先判断节点个数，如果是0-1个，直接返回head，如果大于1个，head->next是要返回的头指针。利用3个指针来做变换操作。p,q用来交换两个指针，r记录上一次交换后的位置，它的next也需要改变。

/** * 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 == NULL || head -> next == NULL)            return head;        ListNode * p = head, *q, *r = new ListNode(0);        head = p -> next;        while(p && (q = p-> next)){            r -> next = q;            p -> next = q -> next;            q -> next = p;            r = p;            p = p -> next;        }        return head;    }};