21. Merge Two Sorted Lists（js)

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->4
Output: 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;
};