LeetCode 24 Swap Nodes in Pairs

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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.

使用遍历。交换前两个节点后，往下两个节点循环前，要记录上个节点哦，否则会丢节点呢。

对于链表操作，别想太多，就是该怎么办就怎么办，凭感觉写就好了。

public ListNode swapPairs(ListNode head) {if (head == null || head.next == null) return head;ListNode result = head.next;ListNode pre = new ListNode(-1);while (head != null && head.next != null) {ListNode even = head.next;head.next = even.next;even.next = head;pre.next = even;pre = head;head = head.next;}return result;}

另外，可以使用递归：

public ListNode swapPairs(ListNode head) {if (head == null || head.next == null) return head;ListNode even = head.next;head.next = swapPairs(even.next);even.next = head;return even;}

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