[24] 两两交换链表中的节点

思路：

1.有一个链表0->1->2->3->4->5->6,创建三个指针，curr，n1,n2分别指向0，1，2

2.交换位置

curr.next=n2即0->2
n1.next=n2.next即1->3
n2.next=n1即2->1
结果就是0->2->1->3->4->5->6

3.再curr=n1也就是新的一轮curr，n1,n2分别指向1，3，4，重复步骤2即可

/*
 * @lc app=leetcode.cn id=24 lang=javascript
 *
 * [24] 两两交换链表中的节点
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var swapPairs = function (head) {
  let dummy = new ListNode()
  dummy.next = head
  let curr = dummy
  while (curr.next !== null && curr.next.next !== null) {
    let n1 = curr.next
    let n2 = curr.next.next
    curr.next = n2
    n1.next = n2.next
    n2.next = n1
    curr = n1
  }
  return dummy.next
}
// @lc code=end

// @after-stub-for-debug-begin
module.exports = swapPairs;
// @after-stub-for-debug-end