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leetcode Merge Two Sorted Lists

代码：

 1 #include<iostream>
 2 
 3 using namespace std;
 4 
 5 struct ListNode {
 6     int val;
 7     ListNode *next;
 8     ListNode(int x) : val(x), next(NULL) {}
 9 };
10 
11 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
12     if (l1 == NULL)
13         return l2;
14     if (l2 == NULL)
15         return l1;
16     ListNode *p = l1, *q = l2;
17     ListNode *head;
18     if (l1->val < l2->val)
19     {
20         head = l1;
21         p = p->next;
22     }
23     else
24     {
25         head = l2;
26         q = q->next;
27     }
28     ListNode *m = head;
29     while (p != NULL&&q != NULL)
30     {
31         if (p->val < q->val)
32         {
33             m->next = p;
34             p = p->next;
35             m = m->next;
36         }
37         else
38         {
39             m->next = q;
40             q = q->next;
41             m = m->next;
42         }
43     }
44     if (p == NULL)
45     {
46         m->next = q;
47     }
48     else
49     {
50         m -> next = p;
51     }
52     return head;
53 }
54 
55 int main()
56 {
57     ListNode *a1 = (ListNode *)malloc(sizeof(ListNode));
58     ListNode *a2 = (ListNode *)malloc(sizeof(ListNode));
59     ListNode *a3 = (ListNode *)malloc(sizeof(ListNode));
60     ListNode *a4 = (ListNode *)malloc(sizeof(ListNode));
61     ListNode *a5 = (ListNode *)malloc(sizeof(ListNode));
62     a1->val = 2;
63     a2->val = 3;
64     a3->val = 6;
65     a4->val = 9;
66     a1->next = NULL;
67     a2->next = a3;
68     a3->next = a4;
69     a4->next = NULL;
70     ListNode *b1 = (ListNode *)malloc(sizeof(ListNode));
71     ListNode *b2 = (ListNode *)malloc(sizeof(ListNode));
72     ListNode *b3 = (ListNode *)malloc(sizeof(ListNode));
73     ListNode *b4 = (ListNode *)malloc(sizeof(ListNode));
74     ListNode *b5 = NULL;
75     b1->val = 1;
76     b2->val = 62;
77     b3->val = 68;
78     b4->val = 90;
79     b1->next = NULL;
80     b2->next = b3;
81     b3->next = b4;
82     b4->next = NULL;
83     ListNode *m = mergeTwoLists(a1, b1);
84     for (; m != NULL; m = m->next)
85         cout << m->val << endl;
86 }

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原文地址：https://www.cnblogs.com/chaiwentao/p/4427398.html