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

分析，拿到lista和listb，从第一个结点开始分析。

对比第一个结点，把值较小的，设置为current1，另外一个设置为current2。

 if (l1 == null)
            {
                return l2;
            }

            if (l2 == null)
            {
                return l1;
            }

            ListNode current1;
            ListNode current2;
            if (l1.val <= l2.val)
            {
                current1 = l1;
                current2 = l2;
            }
            else
            {
                current1 = l2;
                current2 = l1;
            }

 

 

 

  public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
         ListNode current1Prev = null;
            ListNode current2Next;
            ListNode head;
            if (l1 == null)
            {
                return l2;
            }

            if (l2 == null)
            {
                return l1;
            }

            ListNode current1;
            ListNode current2;
            if (l1.val <= l2.val)
            {
                current1 = l1;
                current2 = l2;
            }
            else
            {
                current1 = l2;
                current2 = l1;
            }

            head = current1;
            while (current1 != null && current2 != null)
            {
                if (current1.val <= current2.val)
                {
                    current1Prev = current1;
                    current1 = current1.next;
                }
                else
                {
                    current1Prev.next = current2;
                    current1Prev = current2;
                    current2Next = current2.next;
                    current2.next = current1;
                    current2 = current2Next;
                }
            }

            if (current1 == null)
            {
                current1Prev.next = current2;
            }

            return head;
    }