21. Merge Two Sorted Lists★

题目内容：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.

题目分析：本题是要合并两个已经有序的单链表，思路很简单，有两种方法：非递归和递归。

题目代码：

（1）非递归：

　　为方便操作，定义一个辅助的头节点，然后比较原来两个链表的头节点，将小的那一个加入到合并链表，最后，当其中一个链表为空时，直接将另一个链表接入到合并链表即可。

//public class LeetCode21 为测试
public class LeetCode21 {
    public static void main(String[] args) {
        ListNode m1=new ListNode(1),m2=new ListNode(3),m3=new ListNode(5);
        m1.next=m2;
        m2.next=m3;
        System.out.println("链表1:"+m1.val+"->"+m2.val+"->"+m3.val);
        ListNode n1=new ListNode(2),n2=new ListNode(4),n3=new ListNode(6);
        n1.next=n2;
        n2.next=n3;
        System.out.println("链表2:"+n1.val+"->"+n2.val+"->"+n3.val);
        ListNode result=new Solution().mergeTwoLists(m1, n1);
        if(result!=null){
            System.out.print("合并链表:"+result.val);
            ListNode resultNext=result.next;
            while(resultNext!=null){
                 System.out.print("->"+resultNext.val);
                 resultNext=resultNext.next;
             }
        }
    }
}
class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
         ListNode fakeHead=new ListNode(0);
         ListNode p=fakeHead;
         while(l1!=null&&l2!=null){
             if(l1.val<l2.val){
                 p.next=l1;
                 l1=l1.next;
             }else{
                 p.next=l2;
                 l2=l2.next;
             }
             p=p.next;
         }
         if(l1!=null) p.next=l1;
         if(l2!=null) p.next=l2;
         return fakeHead.next;
        }
}
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
    }

（2）递归：

//public class LeetCode21 为测试
public class LeetCode21 {
    public static void main(String[] args) {
        ListNode m1=new ListNode(1),m2=new ListNode(3),m3=new ListNode(5);
        m1.next=m2;
        m2.next=m3;
        System.out.println("链表1:"+m1.val+"->"+m2.val+"->"+m3.val);
        ListNode n1=new ListNode(2),n2=new ListNode(4),n3=new ListNode(6);
        n1.next=n2;
        n2.next=n3;
        System.out.println("链表2:"+n1.val+"->"+n2.val+"->"+n3.val);
        ListNode result=new Solution().mergeTwoLists(m1, n1);
        if(result!=null){
            System.out.print("合并链表:"+result.val);
            ListNode resultNext=result.next;
            while(resultNext!=null){
                 System.out.print("->"+resultNext.val);
                 resultNext=resultNext.next;
             }
        }
    }
}
class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
         if(l1==null) return l2;
         if(l2==null) return l1;
         if(l1.val<l2.val){
             l1.next=mergeTwoLists(l1.next, l2);
             return l1;
         }else{
             l2.next=mergeTwoLists(l1, l2.next);
             return l2;
         }
        }
}
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
    }