题目：
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.
链接：https://leetcode.com/problems/swap-nodes-in-pairs/#/description
4/12/2017
4ms, 85%
链表题画画图比较好
1 public class Solution { 2 public ListNode swapPairs(ListNode head) { 3 if (head == null || head.next == null) return head; 4 ListNode dummy = new ListNode(-1); 5 dummy.next = head; 6 ListNode prev = dummy; 7 ListNode cur = head; 8 9 while(cur != null && cur.next != null) { 10 prev.next = cur.next; 11 cur.next = cur.next.next; 12 prev.next.next = cur; 13 prev = cur; 14 cur = cur.next; 15 } 16 return dummy.next; 17 } 18 }
别人的recursive做法，虽然空间复杂度高，但是这种思路挺有意思，贴出来的代码比较容易懂一点
https://discuss.leetcode.com/topic/4351/my-accepted-java-code-used-recursion
https://discuss.leetcode.com/topic/41459/java-simple-recursive-solution
1 public class Solution { 2 public ListNode swapPairs(ListNode head) { 3 if (head == null || head.next == null) return head; 4 ListNode second = head.next; 5 ListNode third = second.next; 6 7 second.next = head; 8 head.next = swapPairs(third); 9 10 return second; 11 } 12 }
更多讨论：
https://discuss.leetcode.com/category/32/swap-nodes-in-pairs