标签:amp together return else val nbsp oge nod ret
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.
此题考查链表的操作,比较简单,直接上代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
ListNode dummy = head;
while(l1!=null&&l2!=null){
if(l1.val<l2.val){
ListNode node = new ListNode(l1.val);
l1 = l1.next;
head.next = node;
head = head.next;
}else{
ListNode node = new ListNode(l2.val);
l2 = l2.next;
head.next = node;
head = head.next;
}
}
if(l1!=null){
ListNode node = l1;
head.next = node;
}
if(l2!=null){
ListNode node = l2;
head.next = node;
}
return dummy.next;
}
}
标签:amp together return else val nbsp oge nod ret
原文地址:http://www.cnblogs.com/codeskiller/p/6359905.html