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.
从头遍历链表，将该交换的交换即可：
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var swapPairs = function(head) {
// if (head===null) {
// return head;
// }
// var result = head;
// var pointer = head;
// if (head!==null&&head.next!==null) {
// result = head.next;
// var next = head.next.next;
// head.next.next = head;
// head.next = next;
// }
// var first,second,thrid;
// while (pointer.next!==null&&pointer.next.next!==null) {
// first = pointer.next;
// second = pointer.next.next;
// third = pointer.next.next.next;
// pointer.next = second;
// second.next = first;
// first.next = third;
// pointer = first;
// }
// return result;
var a = new ListNode(0);
a.next = head;
var pointer = a;
while (head&&head.next) {
var third = head.next.next;
pointer.next = head.next;
head.next.next = head;
head.next = third;
pointer = head;
head = third;
}
return a.next;
};
