leetcode 21.合并两个有序链表

解题思路：

创建一个新的链表，然后双指针循环两个有序链表，值小的先合并，值相等都合并，然后再判断如果两个链表有没遍历完的，就直接合并到新链表，最后返回值

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */

function mergeTwoLists (l1, l2) {
        //1.创建头节点
        var head = new ListNode(0)
        var current = head;
        //2.双指针循环遍历
        while(l1 && l2) {
            //2.1 值小的合并到新链表
            if(l1.val < l2.val) {
                current.next = new ListNode(l1.val);
                current = current.next
                l1 = l1.next
            }else if (l1.val == l2.val) { //2.2 值相等都并入
                current.next = new ListNode(l1.val);
                current = current.next;
                l1 = l1.next
                current.next = new ListNode(l2.val);;
                current = current.next;
                l2 = l2.next;
            }else {
                current.next = new ListNode(l2.val);;
                current = current.next;
                l2 = l2.next;
            }
//            console.log(new ListNode())
        }
        //3.检查，没用遍历完的直接并入新链表
        if(l1 != null){
            current.next = l1
        }
        if(l2 != null){
            current.next = l2
        }
        //4.返回head.next
        return head.next
    }
    var l1 = {
        val: 1,
        next: {
            val: 2,
            next: {
                val: 4,
                next: null
            }
        }
    }

    var l2 = {
        val: 1,
        next: {
            val: 3,
            next: {
                val: 5,
                next: {
                    val: 6,
                    next: null
                }
            }
        }
    }
    console.log(mergeTwoLists(l1, l2));

打印结果如下：

 牛客网：

function Merge(pHead1, pHead2)
{
    // write code here
    if(pHead1 == null) {
        return pHead2
    }
    if(pHead2 == null){
        return pHead1
    }
    //定义一个新的head用来合并链表
    let head = new NodeList(0)
    let current = head

    while(pHead1 && pHead2){
        //值小就先合并
        if(pHead1.val < pHead2.val) {
            current.next = pHead1
            current = current.next
            pHead1 = pHead1.next
        }else if(pHead1.val == pHead2.val) { //值相等就都合并
            current.next = pHead1
            current = current.next
            pHead1 = pHead1.next
            current.next = pHead2
            current = current.next
            pHead2 = pHead2.next
        }else {
            current.next = pHead2
            current = current.next
            pHead2 = pHead2.next
        }
    }
    //检查，没有合并完的直接并入
    if(pHead1 !==null){
        current.next = pHead1
    }
    if(pHead2 !==null){
        current.next = pHead2
    }
    return head.next
}