给定两个链接列表，任务是检查第一个列表是否存在于第二个列表中。
例子：
Input : list1 = 10->20
list2 = 5->10->20
Output : LIST FOUND
Input : list1 = 1->2->3->4
list2 = 1->2->1->2->3->4
Output : LIST FOUND
Input : list1 = 1->2->3->4
list2 = 1->2->2->1->2->3
Output : LIST NOT FOUND
算法：
1-取第二个列表的第一个节点。
2-从第一个节点开始匹配第一个列表。
3-如果整个列表匹配，则返回true。
4-否则休息，再次将第一个列表带到第一个节点。
5-将第二个列表带到其第二个节点。
6-重复这些步骤，直到任何链接列表变为空。
7-如果第一个列表为空，则找不到列表。
下面是实现。
C++
// C++ program to find a list in second list
#include
using namespace std;
// A Linked List node
struct Node
{
int data;
Node* next;
};
// Returns true if first list is present in second
// list
bool findList(Node* first, Node* second)
{
Node* ptr1 = first, *ptr2 = second;
// If both linked lists are empty, return true
if (first == NULL && second == NULL)
return true;
// Else If one is empty and other is not return
// false
if ( first == NULL ||
(first != NULL && second == NULL))
return false;
// Traverse the second list by picking nodes
// one by one
while (second != NULL)
{
// Initialize ptr2 with current node of second
ptr2 = second;
// Start matching first list with second list
while (ptr1 != NULL)
{
// If second list becomes empty and first
// not then return false
if (ptr2 == NULL)
return false;
// If data part is same, go to next
// of both lists
else if (ptr1->data == ptr2->data)
{
ptr1 = ptr1->next;
ptr2 = ptr2->next;
}
// If not equal then break the loop
else break;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == NULL)
return true;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second->next;
}
return false;
}
/* Function to print nodes in a given linked list */
void printList(Node* node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
// Function to add new node to linked lists
Node *newNode(int key)
{
Node *temp = new Node;
temp-> data= key;
temp->next = NULL;
return temp;
}
/* Driver program to test above functions*/
int main()
{
/* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node *a = newNode(1);
a->next = newNode(2);
a->next->next = newNode(3);
a->next->next->next = newNode(4);
Node *b = newNode(1);
b->next = newNode(2);
b->next->next = newNode(1);
b->next->next->next = newNode(2);
b->next->next->next->next = newNode(3);
b->next->next->next->next->next = newNode(4);
findList(a,b) ? cout << "LIST FOUND" :
cout << "LIST NOT FOUND";
return 0;
} Java
// Java program to find a list in second list
import java.util.*;
class GFG
{
// A Linked List node
static class Node
{
int data;
Node next;
};
// Returns true if first list is
// present in second list
static boolean findList(Node first,
Node second)
{
Node ptr1 = first, ptr2 = second;
// If both linked lists are empty,
// return true
if (first == null && second == null)
return true;
// Else If one is empty and
// other is not, return false
if (first == null ||
(first != null && second == null))
return false;
// Traverse the second list by
// picking nodes one by one
while (second != null)
{
// Initialize ptr2 with
// current node of second
ptr2 = second;
// Start matching first list
// with second list
while (ptr1 != null)
{
// If second list becomes empty and
// first not then return false
if (ptr2 == null)
return false;
// If data part is same, go to next
// of both lists
else if (ptr1.data == ptr2.data)
{
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
// If not equal then break the loop
else break;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == null)
return true;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second.next;
}
return false;
}
/* Function to print nodes in a given linked list */
static void printList(Node node)
{
while (node != null)
{
System.out.printf("%d ", node.data);
node = node.next;
}
}
// Function to add new node to linked lists
static Node newNode(int key)
{
Node temp = new Node();
temp.data= key;
temp.next = null;
return temp;
}
// Driver Code
public static void main(String[] args)
{
/* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node a = newNode(1);
a.next = newNode(2);
a.next.next = newNode(3);
a.next.next.next = newNode(4);
Node b = newNode(1);
b.next = newNode(2);
b.next.next = newNode(1);
b.next.next.next = newNode(2);
b.next.next.next.next = newNode(3);
b.next.next.next.next.next = newNode(4);
if(findList(a, b) == true)
System.out.println("LIST FOUND");
else
System.out.println("LIST NOT FOUND");
}
}
// This code is contributed by Princi SinghPython3
# Python3 program to find a list in second list
class Node:
def __init__(self, value = 0):
self.value = value
self.next = None
# Returns true if first list is
# present in second list
def findList(first, second):
# If both linked lists are empty/None,
# return True
if not first and not second:
return True
# If ONLY one of them is empty,
# return False
if not first or not second:
return False
ptr1 = first
ptr2 = second
# Traverse the second LL by
# picking nodes one by one
while ptr2:
# Initialize 'ptr2' with current
# node of 'second'
ptr2 = second
# Start matching first LL
# with second LL
while ptr1:
# If second LL become empty and
# first not, return False,
# since first LL has not been
# traversed completely
if not ptr2:
return False
# If value of both nodes from both
# LLs are equal, increment pointers
# for both LLs so that next value
# can be matched
elif ptr1.value == ptr2.value:
ptr1 = ptr1.next
ptr2 = ptr2.next
# If a single mismatch is found
# OR ptr1 is None/empty,break out
# of the while loop and do some checks
else:
break
# check 1 :
# If 'ptr1' is None/empty,that means
# the 'first LL' has been completely
# traversed and matched so return True
if not ptr1:
return True
# If check 1 fails, that means, some
# items for 'first' LL are still yet
# to be matched, so start again by
# bringing back the 'ptr1' to point
# to 1st node of 'first' LL
ptr1 = first
# And increment second node element to next
second = second.next
return False
# Driver Code
# Let us create two linked lists to
# test the above functions.
# Created lists would be be
# node_a: 1->2->3->4
# node_b: 1->2->1->2->3->4
node_a = Node(1)
node_a.next = Node(2)
node_a.next.next = Node(3)
node_a.next.next.next = Node(4)
node_b = Node(1)
node_b.next = Node(2)
node_b.next.next = Node(1)
node_b.next.next.next = Node(2)
node_b.next.next.next.next = Node(3)
node_b.next.next.next.next.next = Node(4)
if findList(node_a, node_b):
print("LIST FOUND")
else:
print("LIST NOT FOUND")
# This code is contributed by GauriShankarBadolaC#
// C# program to find a list in second list
using System;
class GFG
{
// A Linked List node
class Node
{
public int data;
public Node next;
};
// Returns true if first list is
// present in second list
static Boolean findList(Node first,
Node second)
{
Node ptr1 = first, ptr2 = second;
// If both linked lists are empty,
// return true
if (first == null && second == null)
return true;
// Else If one is empty and
// other is not, return false
if (first == null ||
(first != null && second == null))
return false;
// Traverse the second list by
// picking nodes one by one
while (second != null)
{
// Initialize ptr2 with
// current node of second
ptr2 = second;
// Start matching first list
// with second list
while (ptr1 != null)
{
// If second list becomes empty and
// first not then return false
if (ptr2 == null)
return false;
// If data part is same, go to next
// of both lists
else if (ptr1.data == ptr2.data)
{
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
// If not equal then break the loop
else break;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == null)
return true;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second.next;
}
return false;
}
/* Function to print nodes
in a given linked list */
static void printList(Node node)
{
while (node != null)
{
Console.Write("{0} ", node.data);
node = node.next;
}
}
// Function to add new node to linked lists
static Node newNode(int key)
{
Node temp = new Node();
temp.data= key;
temp.next = null;
return temp;
}
// Driver Code
public static void Main(String[] args)
{
/* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node a = newNode(1);
a.next = newNode(2);
a.next.next = newNode(3);
a.next.next.next = newNode(4);
Node b = newNode(1);
b.next = newNode(2);
b.next.next = newNode(1);
b.next.next.next = newNode(2);
b.next.next.next.next = newNode(3);
b.next.next.next.next.next = newNode(4);
if(findList(a, b) == true)
Console.Write("LIST FOUND");
else
Console.Write("LIST NOT FOUND");
}
}
// This code is contributed by Rajput-Ji输出：
LIST FOUND