【牛客题霸题解】合并有序链表

这道题目和合并有序数组得思路是基本上一致的。就是每次选两个链表中的较小值。
注意当一个链表中的元素被选空了以后，剩下都要选择另一个链表里的元素

c++

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        // write code here
        if(l1==NULL)return l2;
        if(l2==NULL)return l1;
        ListNode* ans=new ListNode(-1);
        ListNode* now = ans;
        while( l1!=NULL || l2!=NULL )
        {
            if( l2 == NULL || (l1 != NULL && (l1->val) <= (l2->val)) )
            //l2链表中的数全部用完了就填l1数组中的数 或l1数组中的数没有用完，并且l1数组的数大
            {
                now->next = l1;
                l1 = l1->next;
            }
            else
            {
                now->next = l2;
                l2 = l2->next;
            }
            now = now->next;
        }
        now->next=NULL;
        return ans->next;
    }
};

java

import java.util.*;
public class Solution {
    public ListNode mergeTwoLists (ListNode l1, ListNode l2) {
        // write code here
        if(l1== null)return l2;
        if(l2== null)return l1;
        ListNode ans=new ListNode(-1);
        ListNode now = ans;
        while( l1!=null || l2!= null )
        {
            if( l2 == null || (l1 != null && (l1.val) <= (l2.val)) )
            //l2链表中的数全部用完了就填l1数组中的数 或l1数组中的数没有用完，并且l1数组的数大
            {
                now.next = l1;
                l1 = l1.next;
            }
            else
            {
                now.next = l2;
                l2 = l2.next;
            }
            now = now.next;
        }
        now.next = null;
        return ans.next;
    }
}

python

class Solution:
    def mergeTwoLists(self , l1 , l2 ):
        # write code here
        if l1== None:
            return l2
        if l2== None:
            return l1
        ans = ListNode(-1);
        now = ans;
        while l1!=None or l2!= None:
            if l2 == None or (l1 != None and (l1.val) <= (l2.val)):
            #l2链表中的数全部用完了就填l1数组中的数 或l1数组中的数没有用完，并且l1数组的数大
                now.next = l1
                l1 = l1.next
            else:
                now.next = l2
                l2 = l2.next
            now = now.next;
        now.next = None;
        return ans.next;