题目描述
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

思路1：双指针法

/*** 合并两个有序链表（默认为升序）* 双指针法* @param l1* @param l2* @return*/public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if(l1 == null) {return l2;}if(l2 == null) {return l1;}ListNode head = new ListNode(-1);ListNode index = head;while(l1!=null && l2!=null) {if(l1.val>l2.val) {index.next = l2;l2=l2.next;}else {index.next = l1;l1 = l1.next;}index = index.next;}if(l1!=null) {while (l1!=null) {index.next = l1;l1 = l1.next;index = index.next;}}if(l2!=null) {while (l2!=null) {index.next = l2;l2 = l2.next;index = index.next;}}return head.next;}

思路1代码优化版（代码短了，耗时反而增加了，应该是if else的判断多了导致的）

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if(l1 == null) {return l2;}if(l2 == null) {return l1;}ListNode head = new ListNode(-1);ListNode index = head;while(l1!=null || l2!=null) {if(l1 !=null && l2!=null) {if(l1.val>l2.val) {index.next = l2;l2=l2.next;}else {index.next = l1;l1 = l1.next;}}else if (l1 != null && l2==null) {index.next = l1;l1 = l1.next;}else if (l1 == null && l2!=null) {index.next = l2;l2=l2.next;}index = index.next;}return head.next;}

思路1优化版，优化了时间

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if(l1 == null) {return l2;}if(l2 == null) {return l1;}ListNode head = new ListNode(-1);ListNode index = head;while(l1!=null && l2!=null) {if(l1.val>l2.val) {index.next = l2;l2=l2.next;}else {index.next = l1;l1 = l1.next;}index = index.next;}//任一为空，直接链接另一个链表if(l1!=null) {index.next=l1;}else {index.next=l2;}return head.next;}