LeetCode -- Merge Two sorted lists

题目描述：
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.


就是把两个已经排序的链表进行合并。


思路：
将链表l2合并到l1中，逐个遍历l2
如果l1不是最后一个:
l1<=l2 && l2 <= l1.next  ：移除l2，将l2放在l1.next
l2 < l1 : 将l2.next=l1 并替换头结点
else: l1=l1.next
如果l1是最后结点：
l2 > l1：l1.next = l2
else : 将l2.next = l1并替换头结点




实现代码：

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution {    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {    if(l1 == null){return l2;}if(l2 == null){return l1;}// merge l2 into l1ListNode head = l1;while(l2 != null){if(l1.next != null){if(l2.val >= l1.val && l2.val <= l1.next.val){var t = Remove(ref l2);t.next = l1.next;l1.next = t;}else if(l2.val < l1.val){var t = Remove(ref l2);t.next = l1;head = t;l1 = t;}else{l1 = l1.next;}}else{if(l1.val < l2.val){l1.next = Remove(ref l2);}else{var t = Remove(ref l2);t.next = l1;head = t;l1 = t;}}}return head;    }        private ListNode Remove(ref ListNode n)    {            ListNode t = new ListNode(n.val);    if(n.next == null){    n = null;    }    else{    n.val = n.next.val;    n.next = n.next.next;    }    return t;    }}