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.
使用两个指针，比较第一个链表和第二个链表的元素，如果第二个链表的元素大于第一个的，就把第一个指针后移。否则就将第二个指针所指的元素插在第一个指针的前面。这样做不需要一个新的链表来储存，也比较快。
/**
* 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) {
var pointer1 = l1;
var pointer2 = l2;
while (pointer1!==null&&pointer2!==null) {
if (pointer1.val<=pointer2.val) {
if (pointer1.next===null) {
pointer1.next = pointer2;
return l1;
}
pointer1 = pointer1.next;
} else {
var temp = new ListNode(pointer1.val);
temp.next = pointer1.next;
pointer1.val = pointer2.val;
pointer1.next = temp;
pointer2 = pointer2.next;
}
if (pointer2===null) {
return l1;
}
}
return l1 ? l1 : l2;
};
