leetcode 题解Merge Two Sorted Lists（有序链表归并）

题目：

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.

说明：有序链表归并（从小到大）

1）此链表无头节点

实现：

方法一：非递归

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
         ListNode *la=l1,*lb=l2;
         ListNode *q=NULL,*p=NULL;
         if(la==NULL) return l2;
         if(lb==NULL) return l1;
         //此单链表无头结点，若有头结点，可直接p=head节点即可，无需下面的if...else...
         if((la->val) < (lb->val))
             {
                 p=la;
                 la=la->next;
             }
         else
             {
                 p=lb;
                 lb=lb->next;
             }
         q=p;
         while(la&&lb)
         {
             if((la->val) < (lb->val))
             {
                 p->next=la;
                 p=la;
                 la=la->next;
             }
             else
             {
                 p->next=lb;
                 p=lb;
                 lb=lb->next;
             }
         }
         p->next=la?la:lb;
         return q;
     }
 };

方法二：递归（紫红的泪大牛的代码，博客链接：http://www.cnblogs.com/codingmylife/archive/2012/09/27/2705286.html）

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
         if (l1 == NULL) return l2;
     if (l2 == NULL) return l1;

     ListNode *ret = NULL;

     if (l1->val < l2->val)
     {
         ret = l1;
         ret->next = mergeTwoLists(l1->next, l2);
     }
     else
     {
         ret = l2;
         ret->next = mergeTwoLists(l1, l2->next);
     }

     return ret;
     }
 };