坑爹地多次过，全都是写程序时不注意的小问题，书写习惯还需要进一步改善。遇到的bug有：忘记return语句；定义ListNode runner = head.next，却将判断head==null的情况放在这句之后; 忘记了新的head将不会是原来的那个head，而是head.next；
所以以后遇到runner.next.next的情况要先确保runner.next != null; 遇到runner.next的情况要先确保runner != null
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode swapPairs(ListNode head) { 14 if (head == null) return null; 15 ListNode current = head; 16 ListNode runner = head.next; 17 ListNode prev = new ListNode(0); 18 prev.next = current; 19 if (current != null && runner == null) return current; 20 ListNode newhead = head.next; //save the new head; 21 while (runner.next != null){ 22 if(runner.next.next != null){ 23 current.next = runner.next; 24 runner.next = current; 25 prev.next = runner; 26 current = current.next; 27 runner = runner.next.next.next; 28 prev = prev.next.next; 29 } 30 else break; 31 } 32 current.next = runner.next; 33 runner.next = current; 34 prev.next = runner; 35 return newhead; 36 } 37 }
别人一个很好的解法：
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode swapPairs(ListNode head) { 14 // Start typing your Java solution below 15 // DO NOT write main() function 16 ListNode dummy = new ListNode(0); 17 dummy.next = head; 18 ListNode curr = dummy; 19 ListNode node1 = null; 20 ListNode node2 = null; 21 22 while(curr.next!=null && curr.next.next!=null){ 23 node1 = curr.next; 24 node2 = node1.next; 25 ListNode next =node2.next; 26 curr.next = node2; 27 node2.next = node1; 28 node1.next = next; 29 curr=node1; 30 } 31 return dummy.next; 32 } 33 }
Leetcode: Swap Nodes in Pairs,布布扣,bubuko.com
原文：http://www.cnblogs.com/EdwardLiu/p/3719928.html