算法设计（如何检查两个相同的链表（））

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
C
Java
Python3
C#
C ++
C
Java
Python3
C#

当两个链表具有相同的数据并且数据的排列也相同时, 它们是相同的。例如, 链接列表a(1-> 2-> 3)和b(1-> 2-> 3)相同。。编写一个函数来检查给定的两个链表是否相同。
推荐：请在"实践首先, 在继续解决方案之前。方法1(迭代)
为了确定两个列表是否相同, 我们需要同时遍历两个列表, 并且在遍历时需要比较数据。
C ++

// An iterative C++ program to check if // two linked lists are identical or not #include< bits/stdc++.h> using namespace std; /* Structure for a linked list node */ struct Node { int data; struct Node *next; }; /* Returns true if linked lists a and b are identical, otherwise false */ bool areIdentical( struct Node *a, struct Node *b) { while (a != NULL & & b != NULL) { if (a-> data != b-> data) return false ; /* If we reach here, then a and b are not NULL and their data is same, so move to next nodes in both lists */ a = a-> next; b = b-> next; }// If linked lists are identical, then // 'a' and 'b' must be NULL at this point. return (a == NULL & & b == NULL); }/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push( struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node)); /* put in the data */ new_node-> data = https://www.lsbin.com/new_data; /* link the old list off the new node */ new_node-> next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; }// Driver Code int main() { /* The constructed linked lists are : a: 3-> 2-> 1 b: 3-> 2-> 1 */ struct Node *a = NULL; struct Node *b = NULL; push(& a, 1); push(& a, 2); push(& a, 3); push(& b, 1); push(& b, 2); push(& b, 3); if (areIdentical(a, b)) cout < <"Identical" ; else cout < < "Not identical" ; return 0; }// This code is contributed // by Akanksha Rai

// An iterative C program to check if two linked lists are // identical or not #include< stdio.h> #include< stdlib.h> #include< stdbool.h> /* Structure for a linked list node */ struct Node { int data; struct Node *next; }; /* Returns true if linked lists a and b are identical, otherwise false */ bool areIdentical( struct Node *a, struct Node *b) { while (a != NULL & & b != NULL) { if (a-> data != b-> data) return false ; /* If we reach here, then a and b are not NULL and their data is same, so move to next nodes in both lists */ a = a-> next; b = b-> next; }// If linked lists are identical, then 'a' and 'b' must // be NULL at this point. return (a == NULL & & b == NULL); }/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push( struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node)); /* put in the data*/ new_node-> data= https://www.lsbin.com/new_data; /* link the old list off the new node */ new_node-> next = (*head_ref); /* move the head to point to the new node */ (*head_ref)= new_node; }/* Driver program to test above function */ int main() { /* The constructed linked lists are : a: 3-> 2-> 1 b: 3-> 2-> 1 */ struct Node *a = NULL; struct Node *b = NULL; push(& a, 1); push(& a, 2); push(& a, 3); push(& b, 1); push(& b, 2); push(& b, 3); areIdentical(a, b)? printf ("Identical" ): printf ( "Not identical" ); return 0; }

Java

// An iterative Java program to check if two linked lists // are identical or not class LinkedList { Node head; // head of list/* Linked list Node*/ class Node { int data; Node next; Node( int d) { data = https://www.lsbin.com/d; next = null ; } }/* Returns true if linked lists a and b are identical, otherwise false */ boolean areIdentical(LinkedList listb) { Node a = this .head, b = listb.head; while (a != null & & b != null ) { if (a.data != b.data) return false ; /* If we reach here, then a and b are not null and their data is same, so move to next nodes in both lists */ a = a.next; b = b.next; }// If linked lists are identical, then'a' and 'b' must // be null at this point. return (a == null & & b == null ); }/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /*Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */void push( int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; }/* Driver program to test above functions */ public static void main(String args[]) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); /* The constructed linked lists are : llist1: 3-> 2-> 1 llist2: 3-> 2-> 1 */llist1.push( 1 ); llist1.push( 2 ); llist1.push( 3 ); llist2.push( 1 ); llist2.push( 2 ); llist2.push( 3 ); if (llist1.areIdentical(llist2) == true ) System.out.println( "Identical " ); else System.out.println( "Not identical " ); } } /* This code is contributed by Rajat Mishra */

Python3

# An iterative Java program to check if # two linked lists are identical or not # Linked list Node class Node: def __init__( self , d): self .data = https://www.lsbin.com/d self . next = Noneclass LinkedList: def __init__( self ): self .head = None # head of list # Returns true if linked lists a and b # are identical, otherwise false def areIdentical( self , listb): a = self .head b = listb.head while (a ! = None and b ! = None ): if (a.data ! = b.data): return False# If we reach here, then a and b # are not null and their data is # same, so move to next nodes # in both lists a = a. next b = b. next# If linked lists are identical, # then'a' and 'b' must be null # at this point. return (a = = None and b = = None ) # UTILITY FUNCTIONS TO TEST fun1() and fun2() # Given a reference (pointer to pointer) to the # head of a list and an int, push a new node on # the front of the list. def push( self , new_data):# 1 & 2: Allocate the Node & # Put in the data new_node = Node(new_data) # 3. Make next of new Node as head new_node. next = self .head # 4. Move the head to point to new Node self .head = new_node# Driver Code llist1 = LinkedList() llist2 = LinkedList() # The constructed linked lists are : # llist1: 3-> 2-> 1 # llist2: 3-> 2-> 1 llist1.push( 1 ) llist1.push( 2 ) llist1.push( 3 ) llist2.push( 1 ) llist2.push( 2 ) llist2.push( 3 ) if (llist1.areIdentical(llist2) = = True ): print ( "Identical " ) else : print ( "Not identical " )# This code is contributed by Prerna Saini

// An iterative C# program to // check if two linked lists // are identical or not using System; public class LinkedList { Node head; // head of list/* Linked list Node*/ public class Node { public int data; public Node next; public Node( int d) { data = https://www.lsbin.com/d; next = null ; } }/* Returns true if linked lists a and b are identical, otherwise false */ bool areIdentical(LinkedList listb) { Node a = this .head, b = listb.head; while (a != null & & b != null ) { if (a.data != b.data) return false ; /* If we reach here, then a and b are not null and their data is same, so move to next nodes in both lists */ a = a.next; b = b.next; }// If linked lists are identical, // then'a' and 'b' must // be null at this point. return (a == null & & b == null ); }/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */void push( int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; }/* Driver code */ public static void Main(String []args) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); /* The constructed linked lists are : llist1: 3-> 2-> 1 llist2: 3-> 2-> 1 */llist1.push(1); llist1.push(2); llist1.push(3); llist2.push(1); llist2.push(2); llist2.push(3); if (llist1.areIdentical(llist2) == true ) Console.WriteLine( "Identical " ); else Console.WriteLine( "Not identical " ); } }// This code contributed by Rajput-Ji

输出如下：

Identical

【算法设计（如何检查两个相同的链表（））】方法2(递归)
递归解决方案代码比迭代代码干净得多。但是, 你可能不希望将递归版本用于生产代码, 因为它会使用与列表长度成比例的堆栈空间
C ++

// A recursive C++ function to check if two linked // lists are identical or not bool areIdentical(Node *a, Node *b) { // If both lists are empty if (a == NULL & & b == NULL) return true ; // If both lists are not empty, then data of // current nodes must match, and same should // be recursively true for rest of the nodes. if (a != NULL & & b != NULL) return (a-> data =https://www.lsbin.com/= b-> data) & & areIdentical(a-> next, b-> next); // If we reach here, then one of the lists // is empty and other is not return false ; } //This is code is contributed by rathbhupendra

// A recursive C function to check if two linked // lists are identical or not bool areIdentical( struct Node *a, struct Node *b) { // If both lists are empty if (a == NULL & & b == NULL) return true ; // If both lists are not empty, then data of // current nodes must match, and same should // be recursively true for rest of the nodes. if (a != NULL & & b != NULL) return (a-> data =https://www.lsbin.com/= b-> data) & & areIdentical(a-> next, b-> next); // If we reach here, then one of the lists // is empty and other is not return false ; }

Java

// A recursive Java method to check if two linked // lists are identical or not boolean areIdenticalRecur(Node a, Node b) { // If both lists are empty if (a == null & & b == null ) return true ; // If both lists are not empty, then data of // current nodes must match, and same should // be recursively true for rest of the nodes. if (a != null & & b != null ) return (a.data =https://www.lsbin.com/= b.data) & & areIdenticalRecur(a.next, b.next); // If we reach here, then one of the lists // is empty and other is not return false ; }/* Returns true if linked lists a and b are identical, otherwise false */ boolean areIdentical(LinkedList listb) { return areIdenticalRecur( this .head, listb.head); }

Python3

# A recursive Python3 function to check # if two linked lists are identical or not def areIdentical(a, b): # If both lists are empty if (a = = None and b = = None ): return True# If both lists are not empty, # then data of current nodes must match, # and same should be recursively true # for rest of the nodes. if (a ! = None and b ! = None ): return ((a.data = https://www.lsbin.com/= b.data) and areIdentical(a. next , b. next )) # If we reach here, then one of the lists # is empty and other is not return False# This code is contributed by Princi Singh

// A recursive C# method to // check if two linked lists // are identical or not bool areIdenticalRecur(Node a, Node b) { // If both lists are empty if (a == null & & b == null ) return true ; // If both lists are not empty, then data of // current nodes must match, and same should // be recursively true for rest of the nodes. if (a != null & & b != null ) return (a.data =https://www.lsbin.com/= b.data) & & areIdenticalRecur(a.next, b.next); // If we reach here, then one of the lists // is empty and other is not return false ; } /* Returns true if linked lists a and b are identical, otherwise false */ bool areIdentical(LinkedList listb) { return areIdenticalRecur( this .head, listb.head); } }// This code is contributed by princiraj1992

时间复杂度：迭代和递归版本均为O(n)。 n是a和b中较小列表的长度。
如果你发现上述代码/算法有误, 请写评论, 或者找到解决同一问题的更好方法。