【剑指offer 面试题17】合并两个排序的链表

思路：

　　比较两个链表端点值的大小，通过递归的方式排列。

 #include <iostream>
 using namespace std;

 struct ListNode
 {
     int val;
     ListNode *next;
     ListNode(int v = ):val(v), next(NULL){}
 };

 ListNode *SortedListMerge(ListNode *phead1, ListNode *phead2)
 {
     if(phead1 == NULL)
         return phead2;
     else if(phead2 == NULL)
         return phead1;

     ListNode *pMergedHead = NULL;

     if(phead1->val < phead2->val)
     {
         pMergedHead = phead1;
         pMergedHead->next = SortedListMerge(phead1->next, phead2);
     }
     else
     {
         pMergedHead = phead2;
         pMergedHead->next = SortedListMerge(phead1, phead2->next);
     }

     return pMergedHead;
 }

 int main()
 {

     ListNode *l1 = new ListNode();
     ListNode *h1 = l1;
     for(int i = ; i <= ; i += )
     {
         ListNode *temp = new ListNode(i);
         l1->next = temp;
         l1 = l1->next;
     }

     ListNode *l2 = new ListNode();
     ListNode *h2 = l2;
     for(int i = ; i <= ; i += )
     {
         ListNode *temp = new ListNode(i);
         l2->next = temp;
         l2 = l2->next;
     }

     cout<<"List1: ";
     ListNode *print1 = h1;
     while(print1 != NULL)
     {
         cout<<print1->val<<" ";
         print1 = print1->next;
     }
     cout<<endl;

     cout<<"List2: ";
     ListNode *print2 = h2;
     while(print2 != NULL)
     {
         cout<<print2->val<<" ";
         print2 = print2->next;
     }
     cout<<endl;

     ListNode *mergeList = SortedListMerge(h1, h2);

     cout<<"ListMerge: ";
     ListNode *printlist = mergeList;
     while(printlist != NULL)
     {
         cout<<printlist->val<<" ";
         printlist = printlist->next;
     }
     cout<<endl;

 }

测试结果：

List1:
List2:
ListMerge: