AcWing 36. 合并两个排序的链表
原题链接
简单
作者:
长白秋沙
,
2024-03-29 13:15:35
,
所有人可见
,
阅读 1
合并两个排序的链表
优化前(不设 dummy 节点)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
if(l1 == NULL)
return l2;
if(l2 == NULL)
return l1;
ListNode* head = NULL;
ListNode* p1 = l1;
ListNode* p2 = l2;
if(p1->val < p2->val)
{
head = p1;
p1 = p1->next;
}
else
{
head = p2;
p2 = p2->next;
}
ListNode* p = head;
while(p1 != NULL && p2 != NULL)
{
if(p1->val < p2->val)
{
p->next = p1;
p = p1;
p1 = p1->next;
}
else
{
p->next = p2;
p = p2;
p2 = p2->next;
}
}
if(p1 == NULL)
p->next = p2;
if(p2 == NULL)
p->next = p1;
return head;
}
};
优化后(设 dummy 节点)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(-1);
ListNode* p = dummy;
while(l1 && l2)
{
if(l1->val < l2->val)
{
p->next = l1;
l1 = l1->next;
}
else
{
p->next = l2;
l2 = l2->next;
}
p = p->next;
}
if(l1 == NULL)
p->next = l2;
if(l2 == NULL)
p->next = l1;
return dummy->next;
}
};