leetcode 21. 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.

Seen this question in a real interview before? Yes

简单的归并排序，不说了，直接上代码：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2)
{
if(l1==null && l2==null)
return l1;

ListNode p=l1;
ListNode q=l2;
ListNode head=new ListNode(-1);
ListNode next=head;
// p q 值得是当前的位置上的结点
while(p!=null && q!=null)
{
if(p.val < q.val)
{
next.next=new ListNode(p.val);
next=next.next;
p=p.next;
}
else if(p.val > q.val)
{
next.next=new ListNode(q.val);
next=next.next;
q=q.next;
}else
{
next.next=new ListNode(p.val);
next=next.next;
next.next=new ListNode(q.val);
next=next.next;;
p=p.next;
q=q.next;
}
}

if(p!=null)
next.next=p;

if(q!=null)
next.next=q;

return head.next;
}

public static void main(String[] args)
{
Solution so=new Solution();
so.mergeTwoLists(null, null);
}
}

下面是C++的做法，直接归并即可。

代码如下：

#include <iostream>
#include <stack>
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 = new ListNode(-1);
ListNode* a=head;
while (l1 != NULL && l2 != NULL)
{
if (l1->val < l2->val)
{
a->next = new ListNode(l1->val);
l1 = l1->next;
a = a->next;
}
else
{
a->next = new ListNode(l2->val);
l2 = l2->next;
a = a->next;
}
}

while (l1 != NULL)
{
a->next = new ListNode(l1->val);
l1 = l1->next;
a = a->next;
}

while (l2 != NULL)
{

a->next = new ListNode(l2->val);
l2 = l2->next;
a = a->next;
}
return head->next;
}
};