16.合并两个排序链表——剑指offer

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
 1 class Solution {
 2 public:
 3     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
 4     {
 5         ListNode* pNew = NULL;
 6         ListNode* pCurr = NULL;
 7         if (pHead1 == NULL)
 8         {
 9             return pHead2;
10         }
11         else if (pHead2 == NULL)
12         {
13             return pHead1;
14         }
15         else
16         {
17             while (pHead1 !=NULL && pHead2 !=NULL)
18             {
19                 if (pHead1->val <= pHead2->val)
20                 {
21                     if (pNew == NULL)
22                     {
23                         pNew = pCurr = pHead1;
24                     }
25                     else
26                     {
27                         pCurr->next = pHead1;
28                         pCurr = pCurr->next;
29                     }
30                     pHead1 = pHead1->next;
31                 }
32                 else
33                 {
34                     if (pNew == NULL)
35                     {
36                         pNew = pCurr = pHead2;
37                     }
38                     else
39                     {
40                         pCurr->next = pHead2;
41                         pCurr = pCurr->next;
42                     }
43                     pHead2 = pHead2->next;
44                      
45                 }
46  
47             }
48             if (pHead1==NULL)
49             {
50                 pCurr->next = pHead2;
51             }
52             if (pHead2 == NULL)
53             {
54                 pCurr->next = pHead1;
55             }
56              
57         }
58         return  pNew;
59     }
60 };
61 
62 //递归实现方式：
63  class Solution {
64  public:
65      ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
66      {
67          ListNode* pNew = NULL;
68  
69          if (pHead1 == NULL)
70          {
71              return pHead2;
72          }
73          else if (pHead2 == NULL)
74          {
75              return pHead1;
76          }
77          else
78          {
79              if (pHead1->val <= pHead2->val)
80                  {
81                      pNew = pHead1;
82                      pNew->next = Merge(pHead1->next, pHead2);
83                  }
84                  else
85                  {
86                      pNew = pHead2;
87                      pNew->next = Merge(pHead1, pHead2->next);
88  
89                  }
90          }
91          
92          return  pNew;
93      }
94 };

 

 

posted @ 2019-05-19 16:33  unique_ptr  阅读(135)  评论(0)    收藏  举报