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;
}
};