leetcode 21. Merge Two Sorted Lists-蒲公英云

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;
    }
};