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This article mainly explains "how Linked List-Easy merges the two sorted linked lists". The content in the article is simple and clear, and it is easy to learn and understand. Please follow the editor's train of thought to study and learn "how Linked List-Easy merges the two sorted linked lists".

Merge two sorted linked lists and return a new linked list, which is also sorted.

Ideas for solving the problem:

Create two linked lists, one responsible for saving the header node and the other for recording the results of the comparison.

Language: c

/ * Definition for singly-linked list. * struct ListNode {* int val; * struct ListNode* next; *}; * / struct ListNode* mergeTwoLists (struct ListNode* L1, struct ListNode* 12) {struct ListNode* newlist = (struct ListNode*) malloc (sizeof (struct ListNode)); struct ListNode* temp = (struct ListNode*) malloc (sizeof (struct ListNode)); newlist = temp;while (L1 & & L2) {if (L1-> val)

Val) {temp- > next = L1; L1 = L1-> next;} else {temp- > next = L2; L2 = L2-> next;} temp = temp- > next;} temp- > next = L1? L1: L2: return newlist- > next;}

Language: cpp

/ * 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) {ListNode newlist (INT_MIN); ListNode* temp = & newlist;if (L1 = = NULL & & L2 = NULL) {return NULL } if (L1! = NULL & & L2 = = NULL) {return L1;} if (L1 = = NULL & & L2! = NULL) {return L2;} while (L1 & & L2) {if (L1-> val)

Val) {temp- > next = L1; L1 = L1-> next;} else {temp- > next = L2; L2 = L2-> next;} temp = temp- > next;} temp- > next = L1? L1: L2; return newlist.next;}}

Language:python

# Definition for singly-linked list.# class ListNode (object): # def _ _ init__ (self, x): # self.val = x # self.next = Noneclass Solution (object): def mergeTwoLists (self, L1) L2): ": type L1: ListNode: type L2: ListNode: rtype: ListNode"result = cur = ListNode (0) while L1 and l2:if l1.val < l2.val: cur.next = L1L1 = l1.nextelse: cur.next = L2L2 = l2.next cur = cur .next cur.next = L1 or l2return result.next Thank you for your reading The above is the content of "how Linked List-Easy merges the two sorted linked lists". After the study of this article, I believe you have a deeper understanding of how Linked List-Easy merges the two sorted linked lists, and the specific use needs to be verified in practice. Here is, the editor will push for you more related knowledge points of the article, welcome to follow!

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