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LeetCode: Merge Two Sorted Lists

一点小失误，两次过

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
12         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         ListNode *ret = NULL;
15         ListNode *reti = NULL;
16         if (!l1) return l2;
17         if (!l2) return l1;
18         while (l1 && l2) {
19             if (!ret) {
20                 ret = new ListNode(min(l1->val, l2->val));
21                 reti = ret;
22                 if (l1->val < l2->val) l1 = l1->next;
23                 else l2 = l2->next;
24             }
25             else {
26                 ListNode *tmp = new ListNode(min(l1->val, l2->val));
27                 ret->next = tmp;
28                 ret = ret->next;
29                 if (l1->val < l2->val) l1 = l1->next;
30                 else l2 = l2->next;
31             }
32         }
33         if (!l1) ret->next = l2;
34         if (!l2) ret->next = l1;
35         return reti;
36     }
37 };

下面这段没有造新listnode

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
12         ListNode *ans = NULL;
13         ListNode *p = ans;
14         if (!l1 && !l2) return ans;
15         while (l1 || l2) {
16             if (!l1) {
17                 if (!ans) return l2;
18                 p->next = l2;
19                 break;
20             }
21             if (!l2) {
22                 if (!ans) return l1;
23                 p->next = l1;
24                 break;
25             }
26             if (l1->val < l2->val) {
27                 if (!ans) {
28                     ans = l1;
29                     p = ans;
30                 }
31                 else {
32                     p->next = l1;
33                     p = p->next;
34                 }
35                 l1 = l1->next;
36             }
37             else {
38                 if (!ans) {
39                     ans = l2;
40                     p = ans;
41                 }
42                 else {
43                     p->next = l2;
44                     p = p->next;
45                 }
46                 l2 = l2->next;
47             }
48         }
49         return ans;
50     }
51 };

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public int val;
 5  *     public ListNode next;
 6  *     public ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
11         ListNode ans = null;
12         ListNode p = ans;
13         while (l1 != null || l2 != null) {
14             if (l1 == null) {
15                 if (ans == null) return l2;
16                 p.next = l2;
17                 break;
18             }
19             if (l2 == null) {
20                 if (ans == null) return l1;
21                 p.next = l1;
22                 break;
23             }
24             if (l1.val < l2.val) {
25                 if (ans == null) p = ans = l1;
26                 else {
27                     p.next = l1;
28                     p = p.next;
29                 }
30                 l1 = l1.next;
31             }
32             else {
33                 if (ans == null) p = ans = l2;
34                 else {
35                     p.next = l2;
36                     p = p.next;
37                 }
38                 l2 = l2.next;
39             }
40         }
41         return ans;
42     }
43 }

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原文地址：https://www.cnblogs.com/yingzhongwen/p/3010591.html