21. Merge Two Sorted Lists
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.
Example:
Input: 1->2->4, 1->3->4Output: 1->1->2->3->4->4
题意：按照大小顺序合并两个链表
代码如下：
/** * Definition for singly-linked list. * function ListNode(val) { * ????this.val = val; * ????this.next = null; * } *//** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */var mergeTwoLists = function(l1, l2) { ????//判断两链表是否为空 ???????if(!l1 || !l2) return l1===null?l2:l1; ???????if(!l1 && !l2) return null; ???????????????//创建新链表存储结果, ???????let head; ???????//创建指针分别沿着l1,l2遍历 ???????let node1=l1,node2=l2; ???????//比较l1,l2头结点，较小的赋给新链表 ???????if(l1.val>l2.val){ ???????????head=l2; ???????????node2=node2.next; ???????}else{ ???????????head=l1; ???????????node1=node1.next; ???????} ???????//创建指针沿着新链表移动 ???????let node=head; ???????//不断比较两链表对应节点，直到其中一链表到尾 ???????while(node1 && node2){ ???????????if(node1.val>node2.val){ ???????????????node.next=node2; ???????????????node=node.next; ???????????????node2=node2.next; ???????????}else{ ???????????????node.next=node1; ???????????????node=node.next; ???????????????node1=node1.next; ???????????} ???????} ???????????????//判断两链表谁先到头 ???????if(!node1 && !node2){ ???????????return head; ???????} ???????if(!node2&&node1){ ???????????node.next=node1; ???????} ???????if(!node1&& node2){ ???????????node.next=node2; ???????} ???????????????return head;};21. Merge Two Sorted Lists（js)
原文地址：https://www.cnblogs.com/xingguozhiming/p/10387366.html