题目：
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
解决方案：276ms
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 mergeTwoLists(ListNode l1, ListNode l2) { 14 ListNode head = new ListNode(0); 15 ListNode newList = head; 16 17 if(l1 == null || l2 == null){//两者有为空的情况 18 if(l1==null) 19 return l2; 20 if(l2 == null) 21 return l1; 22 } 23 24 while(l1!=null && l2!=null){//两者都不为空 25 26 if(l1.val <= l2.val){ 27 newList.next = l1; 28 l1 = l1.next; 29 }else{// if(l1.val < l2.val) 30 newList.next = l2; 31 l2 = l2.next; 32 } 33 newList = newList.next; 34 } 35 while(l1!=null){ 36 newList.next = l1; 37 l1 = l1.next; 38 newList = newList.next; 39 } 40 while(l2!=null){ 41 newList.next = l2; 42 l2 = l2.next; 43 newList = newList.next; 44 } 45 46 return head.next; 47 } 48 }
总结：
这道题目很简单，就是我们以前数据结构学习的时候的一道课本上的算法代码，其实思路就在那里，很容易就想到，但是我这里犯错了，我在第一个while中把nextList = nextList.next放到了while第一句里面，当然下面的代码也就是nextList = l1;这样了，但是这样是不对的，具体原因我也没弄清楚，希望有人帮我解决问题。多谢诸位。
贴错误代码：
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode head = new ListNode(0); ListNode newList = head; if(l1 == null || l2 == null){//两者有为空的情况 if(l1==null) return l2; if(l2 == null) return l1; } while(l1!=null && l2!=null){//两者都不为空 newList = newList.next; if(l1.val <= l2.val){ newList = l1; l1 = l1.next; }else{// if(l1.val < l2.val) newList = l2; l2 = l2.next; } } while(l1!=null){ newList.next = l1; l1 = l1.next; newList = newList.next; } while(l2!=null){ newList.next = l2; l2 = l2.next; newList = newList.next; } return head.next; } }
代码提示错误：
Input: {2}, {1}
Output: {}
Expected: {1,2}
纠结，这个问题都没搞懂。。。