21. 合并两个有序链表
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.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
方法1
递归解决方案
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == nullptr)
return l2;
if(l2 == nullptr)
return l1;
if(l1->val <= l2->val)
{
l1->next = mergeTwoLists(l1->next,l2);
return l1;
}
else
{
l2->next = mergeTwoLists(l1,l2->next);
returnl2;
}
}
};
方法2
非递归不使用辅助头结点方案
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == nullptr)
return l2;
if(l2 == nullptr)
return l1;
ListNode *head = nullptr;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *last = nullptr;
while(p1!=nullptr && p2!=nullptr)
{
if(p1->val <= p2->val)
{
if(head == nullptr)
{
head = p1;
last = head;
}
else
{
last->next = p1;
last = p1;
}
p1 = p1->next;
}
else
{
if(head == nullptr)
{
head = p2;
last = head;
}
else
{
last->next = p2;
last = p2;
}
p2 = p2->next;
}
}
if (p1 != nullptr)
last->next = p1;
if (p2 != nullptr)
last->next = p2;
return head;
}
};
方法3
非递归使用辅助头结点方案
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == nullptr)
return l2;
if(l2 == nullptr)
return l1;
ListNode *dummy = new ListNode(-1);
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *last = dummy;
while(p1!=nullptr && p2!=nullptr)
{
if(p1->val <= p2->val)
{
last->next = p1;
last = p1;
p1 = p1->next;
}
else
{
last->next = p2;
last = p2;
p2 = p2->next;
}
}
if (p1 != nullptr)
last->next = p1;
if (p2 != nullptr)
last->next = p2;
ListNode *ret = dummy->next;
dummy->next = nullptr;
delete dummy;
return ret;
}
};
