https://leetcode.com/problems/merge-two-sorted-lists/#/description

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.

覃超：clarification:問是不是sorted
寫程序第一步永遠永遠先判斷valid parameters

Mar 21,2017 Version:
Today I wrote this question again. This time I understand ** fakehead ** better.
The key is to think clearly about the while clause and the '&&' symbol in it.

    /**
     * Assume there's a testcase:
     * l1 : 1->3->5->7
     * l2 : 2->4
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode res = new ListNode(-1);
        ListNode fakeHead = res;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                res.next = l1;
                res = res.next;
                l1 = l1.next;
            } else {
                res.next = l2;
                res = res.next;
                l2 = l2.next;
            }
        }
        if (l1 != null) {
            res.next = l1;
        } else {
            res.next = l2;
        }
        return fakeHead.next;
    }

I also found a recursive version in the solutions:

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        
        ListNode mergeHead;
        if(l1.val < l2.val){
            mergeHead = l1;
            mergeHead.next = mergeTwoLists(l1.next, l2);
        }
        else{
            mergeHead = l2;
            mergeHead.next = mergeTwoLists(l1, l2.next);
        }
        return mergeHead;
    }

Mar1 Version:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode fakeHead = new ListNode(0);
        //pointer是fakeHead對象的引用忽媒，以后的操作都在fakehead上
        ListNode pointer = fakeHead;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                pointer.next = l1;
                l1 = l1.next;
            } else {
                pointer.next = l2;
                l2 = l2.next;
            }
            //注意這一行蟹漓，如果不加，pointer就一直停留在原地了
            pointer = pointer.next;
        }
        
        if (l1 != null) {
            pointer.next = l1;
        } else {
            pointer.next = l2;
        }
        return fakeHead.next;
    }