本文概述
编写一个程序, 检查给定的链表是否包含循环, 如果存在循环, 则返回循环中的节点数。例如, 在下面链接的列表中存在一个循环, 并且循环的长度为4。如果不存在该循环, 则该函数应返回0。
方法:在这篇文章中, 我们将使用地图存储存在的节点的地址链表作为键, 其位置作为值。
以下是分步方法:
- 遍历链表的每个节点, 并保持位置从一个开始。在每个节点之后增加位置。
- 检查地图中是否存在该节点。
- 如果地图不包含该节点的地址, 则将其及其位置插入地图。
- 如果地图已经包含该节点的地址, 则返回其位置之间的差。
- 如果找不到这样的节点, 则返回0。
下面是上述方法的实现:
C ++
//C++ program to find length of loop
//in a linked list using Map
#include <bits/stdc++.h>
using namespace std;
//Linked List node
struct Node {
int data;
struct Node* next;
Node( int num)
{
data = num;
next = NULL;
}
};
//Function detects and counts loop
//nodes in the list. If loop is not there, //then returns 0
int countNodesinLoop( struct Node* head)
{
struct Node* p = head;
int pos = 0;
//Maintain a map to store addresses
//of node and their position
unordered_map<Node*, int> m;
//Traverse through the linked list
while (p != NULL) {
//If the node is not present in the map
if (m.find(p) == m.end()) {
m = pos;
pos++;
}
//if the node is present
else {
//Return difference between
//position of the present node and
//position where that node occured before
return (pos - m
);
}
p = p->next;
}
//Return 0 to indicate
//there is no loop
return 0;
}
//Driver code
int main()
{
//Create nodes of the linked list
struct Node* head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);
//Create a loop for testing the function
head->next->next->next->next->next = head->next;
//Call the function for the above linked list
cout <<countNodesinLoop(head) <<endl;
return 0;
}
Java
//Java program to find length of loop
//in a linked list using Map
import java.util.*;
import java.io.*;
class GFG{
static class Node
{
int data;
Node next;
//Constructor
Node( int num)
{
data = num;
next = null ;
}
}
//Function detects and counts loop
//nodes in the list. If loop is not there, //then returns 0
public static int countNodesinLoop(Node head)
{
Node p = head;
int pos = 0 ;
//Maintain a map to store addresses
//of node and their position
HashMap<Node, Integer> m = new HashMap<Node, Integer>();
//Traverse through the linked list
while (p != null )
{
//If the node is not present in the map
if (!m.containsKey(p))
{
m.put(p, pos);
pos++;
}
//If the node is present
else
{
//Return difference between
//position of the present
//node and position where
//that node occured before
return (pos - m.get(p));
}
p = p.next;
}
//Return 0 to indicate
//there is no loop
return 0 ;
}
//Driver code
public static void main (String[] args)
{
//Create nodes of the linked list
Node head = new Node( 1 );
head.next = new Node( 2 );
head.next.next = new Node( 3 );
head.next.next.next = new Node( 4 );
head.next.next.next.next = new Node( 5 );
//Create a loop for testing the function
head.next.next.next.next.next = head.next;
//Call the function for the above linked list
System.out.println(countNodesinLoop(head));
}
}
//This code is contributed by adityapande88
输出如下
4
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