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。
示例1
输入
{1,3,5},{2,4,6}
返回值
{1,2,3,4,5,6}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
| /*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
ListNode* result = NULL;
ListNode* cur = NULL;
if(!pHead1)
return pHead2;
if(!pHead2)
return pHead1;
while(pHead1||pHead2)
{
if(pHead1&&pHead2)
{ if(pHead1->val>=pHead2->val)
{
ListNode* temp = pHead2->next;
if(!result)
{
result = pHead2;
cur = result;
}
else
{
result->next = pHead2;
result = result->next;
}
pHead2 = temp;
}
else
{
ListNode* temp = pHead1->next;
if(!result)
{
result = pHead1;
cur = result;
}
else
{
result->next = pHead1;
result = result->next;
}
pHead1 = temp;
}
}
else
{
if(pHead1)
{
result->next = pHead1;
}
else
{
result->next = pHead2;
}
break;
}
}
return cur;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
| /*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
ListNode* result = new ListNode(-1);
ListNode* cur = result;
if(!pHead1)
return pHead2;
if(!pHead2)
return pHead1;
while(pHead1||pHead2)
{
if(pHead1&&pHead2)
{ if(pHead1->val>=pHead2->val)
{
result->next = pHead2;
pHead2 = pHead2->next;
}
else
{
result->next = pHead1;
pHead1 = pHead1->next;
}
result = result->next;
}
else
{
result->next = pHead1?pHead1:pHead2;
break;
}
}
return cur->next;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
| /*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
if(!pHead1)
return pHead2;
if(!pHead2)
return pHead1;
if(pHead1->val>=pHead2->val)
{
pHead2->next = Merge(pHead1,pHead2->next);
return pHead2;
}else
{
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}
}
};
|