Leet Code OJ 21. Merge Two Sorted Lists [Difficulty: Easy]

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.

翻译:
合并2个已经排序的链表，并且返回一个新的链表。这个新的链表应该由前面提到的2个链表的节点所组成。

分析：
注意头节点的处理，和链表结束（next为null）的处理。以下代码新增了一个头指针，来把头节点的处理和普通节点的处理统一了。

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 currentNode=head;while(true){if(l1==null&&l2==null){break;}else if(l2!=null&&(l1==null||l1.val>l2.val)){currentNode.next=l2;l2=l2.next;}else{currentNode.next=l1;l1=l1.next;}currentNode=currentNode.next;}return head.next;}
}

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