python merge two lists

1/**2* Definition for singly-linked list.3* Function ListNode (val) {4* This.val = val;5* This.next = null;6* }7*/8/**9* @param {ListNode} l110* @param {ListNode} L211* @return {ListNode}12*/13var mergetwolists =function(L1, L2) {14var ret =New ListNode (0),p =Ret1617while (L1!==Null L2!==Null) {18if (L1.val L2.val) {P.next =L1;L1 =L1.next;21}Else{P.next =L2;L2 =L2.next;24}25p =P.next;27}2829if (L1!==Null) {P.next =L1;31}3233if (L2!==null) {p.next = L2; The ret.next of the Panax Notoginseng re

1listnode* mergetwolists (listnode* L1, listnode*L2) {2listnode*T;3t= (listnode*) malloc (sizeof(ListNode));4ListNode *T1;5t1=T;6 if(l1==nulll2==NULL)7 returnNULL;8 while(l1L2)9 {Ten if(l1->valval) One { At1->next=L1; -L1=l1->Next; - } the Else - { -t1->next=L2; -L2=l2->Next; + } -T1=t1->Next; + } A if(l1==NULL) at { -t1->next=L2; - } - if(l2==NULL)

C + + code/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Mergetwolists (listnode* L1, listnode*L2) { if(nullptr = = L1)returnL2; if(nullptr = = L2)returnL1; ListNode*ret; if(L1->val val) {ret=L1; L1= l1->Next; } Else{ret=L2; L2= l2->Next; } ListNode* cur =ret; while(nullptr! = L1 nullptr! =L2) { if(L1->val val) {cur->next =L1; L1= l1->Next; } Else{cur->ne

1, linked list definitiontypedef struct LISTELEMENT_T_ { void *data; struct Listelement_t_ *next;} listelement_t;typedef struct list_t_{ int size; int capacity; listelement_t *head; listelement_t *tail;} list_t;2. Merging two ordered linked listslistelement_t * Mergelist (listelement_t *phead1, listelement_t *phead2, int (*compare) (const void *key1, const void *key2 ) {if (compare = = NULL) return null; if (PHead1 = = NULL) return pHead2; if (pHead2 = = NULL) return p

If you think about it, even if you have helplessness and regret, you will feel that happiness cannot be achieved. Should you be thankful? From then on, there will be an instinctive resistance to the incomplete completion. /** This Program Merge two non-descending sequence linked lists. * The values of member variables of the elements in the merged linked list are also in non-descending order. */# Includ

Question Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Method Use the thought of merging and sorting to merge the data until it finally becomes one. public ListNode mergeKLists(ArrayList lists) { int len = lists.size(); if(len == 0){

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Ideas. First of all, if there is an empty list, if there is, then directly return to another linked list, if not, then start comparing the current node of the two linked list, return the smaller elements as precursors, and the pointer moves backward one bit, and then co

Title Description: (link)Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Problem Solving Ideas: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,

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Test instructions: Merging two sorted listsThe problem is not difficult, but there are a lot of details that need attention, direct sticker code1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * struct ListNode *next;6 * };7 */8 structlist

Summary: Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Java:/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode mergetwolists (listnode L1, ListNode L2) {ListNode P1=L1; ListNode P2=L2; ListNode Fakehead=

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Solution: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 * }Ten * } One */ A Public classSolution { - PublicListNode mergetwolists (listnode L1, ListNode

https://leetcode.com/problems/merge-two-sorted-lists/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 *temp; -ListNode * tail=NULL; -ListNode * res=NULL; theListNode * head1=L1; -ListNode * head2=L2; -

https://leetcode.com/problems/merge-two-sorted-lists/classSolution:#@param {ListNode} L1 #@param {ListNode} L2 #@return {ListNode} defmergetwolists (self, L1, L2):ifL1 isNone:returnL2ifL2 isNone:returnL1ifL1.val >L2.val:l1, L2=L2, L1 l1.next=self.mergetwolists (L1.next, L2)returnL1Before abruptly the test instructions understand to be sure to create a new list and return, I'm not happy.Just look at

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Idea: using pseudo-headclassSolution { Public: ListNode*mergetwolists (ListNode *l1, ListNode *L2) {ListNode Fakehead (0); ListNode*P1 =L1; ListNode*P2 =L2; ListNode*pnew = Fakehead; while(P1! = NULL P2! =NULL) { if(P1->val val) {pnew->next =P1; P1= p1->Nex

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Hide TagsLinked List/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*mergetwolists (ListNode *l1, ListNode *L2) { if(l1==NULL)returnL2; if(l2==

This topic is meant to synthesize an ordered list of two ordered lists, and to investigate the merging algorithm and the operation of the linked list.The code is relatively simple, simply say the function of the three pointers in the merge function, sum is the first pointer to return, CUR is the link to return to the location, put is to take the L1 or L2 in a certain pointer node. The full operating code is

https://oj.leetcode.com/problems/merge-two-sorted-lists/http://blog.csdn.net/linhuanmars/article/details/19712593/***definitionforsingly-linkedlist.*publicclasslistnode {*intval;*ListNode Next;*listnode (intx) {* val=x;*next =null;*}*}*/publicclasssolution {publiclistnodemergetwolists (LISTNODENBSP;L1,NBSP;LISTNODENBSP;L2 ) { if (l1==nulll2== NULL) returnnull; listnodetoreturn=null; ListNodecurNode=null; Li

Merges two incrementally linked lists into an incremented new list.Idea: Compare the head node size of two increments of the linked list, assuming that the value of the head node of the list 1 is small, then the head node of the list 1 is the head node of the new linked list, then continue to merge, then compare the second node of the list 1 with the head node of the linked list 2, two linked list or sequen

Merge k Sorted Lists Merge K sorted linked lists and return it as one sorted list. Analyze and describe its complexity. In fact, there is no "skill" in this question, but it is not good to think more. It's just two ways: 1. The number of the first number of columns to form an array, and then take the largest, insert a

https://oj.leetcode.com/problems/merge-k-sorted-lists/ Merges K-sorted arrays and analyzes the complexity of the entire algorithm. Problem-Solving report: First, regardless of any optimization, the most simple way to achieve is to traverse the listAlgorithm complexity is O (KN) public class Solution {ListNode mergetwolists (ListNode list1, ListNode list2) {listnode head = new Listno De (-1);