python merge two lists

translation合并两个排好序的链表，并返回这个新链表。新链表应该由这两个链表的头部拼接而成。Originaltwoandreturnitasanewnewbytheofthefirsttwo lists.Code/** * Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x ): Val (x), Next (NULL) {}}; */ class solution {public:listnode* mergetwolists (listnode* L1, listnode* L2) {if(L2 = = NULL) {returnL1; }if(L1 = = NULL) {returnL2; }if(l1->Val> l2->Val) {listnode* temp = L2; Temp->next = mergetwolists (L1, L2->next);returnTemp }Else{listnode* temp =

1 classSolution {2 Public:3listnode* mergetwolists (listnode* L1, listnode*L2) {4listnode* head =NewListNode (0);5listnode* res =head;6 while(L1! = NULL L2! =NULL) {7 if(L1->val val) {8Head->next =L1;9Head = head->Next;TenL1 = l1->Next; One } A Else{ -Head->next =L2; -Head = head->Next; theL2 = l2->Next; - } - } - if(L1! = NULL) Head->next =L1; + Else if(L2! = NULL) Head->next =L2; - returnRes->Next; +

Iterative:1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7 * };8 */9 classSolution {Ten Public: OneListNode *mergetwolists (ListNode *l1, ListNode *L2) { A if(!L1)returnL2; - if(!L2)returnL1; -ListNode *result =NewListNode (0); theListNode *runner =result; - while(L1 | |L2) { - intTMP1 = Int_max, TMP2 =Int_max; - if(L1) { +TMP1 = l1->Val; -

Class solution{ Public: ListNode*Mergetwolists (ListNode*L1, ListNode*L2) {if(L1== NULL) {returnL2; }if(L2== NULL) {returnL1; } ListNode*Help= NewListNode (0); ListNode*Head=Help while(L1L2) {if(L1 -Val>L2 -val) {Help -Next=L2; L2=L2 -Next }Else{Help -Next=L1; L1=L1 -Next } Help=Help -Next }if(L1) {Help -Next=L1; }if(L2) {Help -Next=L2; }returnHead -Next }};A recursive solution to the Web:ListNode*Mergetwolists (ListNode*L1, ListNode*L2) {if(L1== NULL)returnL2;if(L2== NULL)retu

1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7 * };8 */9 classSolution {Ten Public: Onelistnode* mergetwolists (listnode* L1, listnode*L2) { AListNode *l3 = NULL;//Head Pointer -ListNode *p =L3; - //if (L1 = = NULL L2 = = null) the //return NULL; - //else if - //return (L1 = = NULL)? l1::l2; - if(L1 = = NULL)returnL2; + if(L2 = = NULL)returnL1; -

Label: style blog Io color for SP Div on Log class Solution { public: int listSize(ListNode* l){ if(l==NULL) return 0; int len=1; ListNode* p=l; while (p->next) { p=p->next; ++len; } return len; } ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1==NULL) return l2; if(l2==NULL) return l1; int len1=listSize(l1);int len2=listSize(l2); ListN

1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public ListNode mergeTwoLists(ListNode l1, ListNode l2) {14 if(l1==null) return l2;15 if(l2==null) return l1;16 ListNode head=l1.val Merge two sorted

I wrote it again, and the code looked much refreshed. 1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {12 ListNode tmp(-1);13 ListNode *resHead = tmp;14 ListNode *resHead_output = resHead;15 while(l1 != NULL l2 != NULL){16 if(l1->val Leet

This chapter talks about using for and range to modify the list, and most people will think of the following code when they encounter this scenario:But why the result is not right?Let's take a look at its running steps:1. Remove an object from List A2. Assign a value to X for the object you just made3.x self-Add 1As you can see here, x refers to Object 1, which is the same object as the object referenced by the first object in the list, and it should be replaced, the problem is here, in fact, th

ll和lr都会接到一个元素的列表 # before the last recursion the LL and RL will receive the ordered subsequence LL = Merge_sort (left) RL =merge_sort (right) # We return the two split results after sorting and returning the correct order of the sub-list # Here we call to carry a function to help us in order to merge LL and LR return merge (LL, RL) #这里接收两个列表def merge (left, ri

join and specify Keys (row index) \ r \ n ', concat ([df1,df2],keys=[' A ', ' B ']) # Here are the duplicate data print ' go back \ r \ n ', concat ([df1,df2],ignore_index=true). Drop_duplicates ()The output is:Internal connection by Axis City rank City rank0 Chicago 1 Chicago San Francisco 2 Boston New York City 3 Los Angeles 5 outer Joins and assign keys (row index) City Ranka 0 Chicago 1 1 San Francisco 2 2 New York City 3b 0

Simple learning notes for lists, dictionaries, and tuples in Python, and python learning notes List List is the most flexible ordered collection object type in Python. Different from a string, a list can contain any type of objects: Numbers, strings, or even other lists. A l

Python programming to achieve Merge Sorting, python Merge Sorting Because a question about leetcode last week (Median of Two Sorted Arrays), I would like to carefully understand the implementation of Merge Sorting. I 'd like to explain the sorting Logic first: First of all,

Three methods of String concatenation in Python: python merge and merge In Python, we often encounter the problem of String concatenation. Here I summarize the following three String concatenation methods: 1. Splice with the plus sign (+) The plus sign (+) concatenation is t

Python Learning notes 2--python file types, variables, values, strings, tuples, lists, dictionariesOne, Python file type1. Source CodePython source code file with a. py extension, interpreted by the Pyton program, does not require compilation[[email protected] day01]# vim 1.py#!/usr/bin/

If an element exists in the list, the returned Task:You have a list L, there is an index number I, if I is a valid index, return l[i], if not, return the default value VSolution: The list supports bidirectional indexing, so I can be negative >>> def list_get (L,i,v=none): if -len (l) Len (l): return L[i] else : return v >>> list_get ([1,2,3,4,5,6],3 4 Exception mechanismdef list2_get (l,i,v=None): try: return l[i] except indexerror: return '

[Python] More Pythonic lists are merged with the Amway and pythonpythonic of Python. Original question: https://segmentfault.com/q/1010000005904259 Question: How can I combine a list like L = [1, 2, 3], [4, 5, 6], [7, 8, 9] into a list like [1, 2, 3, 4, 5, 6, 7, 8, 9? The most direct method (By: pine forest) ret = []for x in L: ret += xprint(x) Use the itertoo

Example of how to sort tuples and lists by conditions in Python. Example of python Sort a single tuple in python My colleague Axel Hecht showed me something I don't know about python sorting. In python, you can sort a single tuple

The case of merging multiple lists into a list is encountered. Use the chain method to merge multiple lists. It can also be used to merge Django's QuerySet.1. Python uses chain to merge multiple

Python uses case-insensitive sorting to sort string lists. python uses case-insensitive sorting. This article describes how to sort string lists by case-insensitive python. Share it with you for your reference. The specific analysis is as follows: Let's take a look at the fo