Leetcode21. Merge Two Sorted Lists

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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

这题断断续续刷了好久，好讨厌coding被人打断，忧桑...

其实还挺简单的一道链表合并。

#include<iostream>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* head = NULL;
if (l1 == NULL&&l2 == NULL)
return 0;
else if (l1 == NULL&&l2 != NULL)
return l1;
else if (l1 != NULL&&l2 == NULL)
return l2;
ListNode* p = l1;
ListNode* q = l2;
ListNode* tail = NULL;
if (p->val <= q->val)
{
head = p;
p = p->next;
}
else
{
head = q;
q = q->next;
}
tail = head;
while (p != NULL&&q != NULL)
{
if (p->val <= q->val)
{
tail->next = p;
p = p->next;
}
else
{
tail->next = q;
q = q->next;
}
tail = tail->next;

}
if (p != NULL)
tail->next = p;
else
tail->next = q;
return head;

}
};
int main()
{
ListNode* a = new ListNode(1);
ListNode* b = new ListNode(2);
ListNode* c = new ListNode(3);
ListNode* d = new ListNode(1);
ListNode* e = new ListNode(3);
ListNode* f = new ListNode(4);
a = NULL;
d = NULL;
/*a->next = b;
b->next = c;
c->next = NULL;
d->next = e;
e->next = f;
f->next = NULL;*/
Solution s;
ListNode* head;
head = s.mergeTwoLists(a, d);
while (head != NULL)
{
cout << head->val;
head = head->next;
}
}