题目描述:
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.
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 */
递归解决方案：
1 class Solution {
2 public:
3 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
4
5 if(l1 == nullptr)
6 return l2;
7 if(l2 == nullptr)
8 return l1;
9 if(l1->val <= l2->val)
10 {
11 l1->next = mergeTwoLists(l1->next,l2);
12 return l1;
13 }
14 else
15 {
16 l2->next = mergeTwoLists(l1,l2->next);
17 returnl2;
18 }
19
20 }
21 };
非递归不使用辅助头结点方案：
1 class Solution {
2 public:
3 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
4
5 if(l1 == nullptr)
6 return l2;
7 if(l2 == nullptr)
8 return l1;
9
10 ListNode *head = nullptr;
11 ListNode *p1 = l1;
12 ListNode *p2 = l2;
13 ListNode *last = nullptr;
14 while(p1!=nullptr && p2!=nullptr)
15 {
16 if(p1->val <= p2->val)
17 {
18 if(head == nullptr)
19 {
20 head = p1;
21 last = head;
22 }
23 else
24 {
25 last->next = p1;
26 last = p1;
27 }
28
29 p1 = p1->next;
30 }
31 else
32 {
33 if(head == nullptr)
34 {
35 head = p2;
36 last = head;
37 }
38 else
39 {
40 last->next = p2;
41 last = p2;
42 }
43
44 p2 = p2->next;
45 }
46 }
47
48 if (p1 != nullptr)
49 last->next = p1;
50 if (p2 != nullptr)
51 last->next = p2;
52
53 return head;
54
55 }
56 };
非递归使用辅助头结点方案：
1 class Solution {
2 public:
3 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
4
5 if(l1 == nullptr)
6 return l2;
7 if(l2 == nullptr)
8 return l1;
9
10 ListNode *dummy = new ListNode(-1);
11 ListNode *p1 = l1;
12 ListNode *p2 = l2;
13 ListNode *last = dummy;
14 while(p1!=nullptr && p2!=nullptr)
15 {
16 if(p1->val <= p2->val)
17 {
18 last->next = p1;
19 last = p1;
20 p1 = p1->next;
21 }
22 else
23 {
24 last->next = p2;
25 last = p2;
26 p2 = p2->next;
27 }
28 }
29
30 if (p1 != nullptr)
31 last->next = p1;
32 if (p2 != nullptr)
33 last->next = p2;
34
35 ListNode *ret = dummy->next;
36 dummy->next = nullptr;
37 delete dummy;
38
39 return ret;
40
41 }
42 };