给定一个单链表。任务是检查链接列表中连续节点之间的绝对差是否为1。
例子：
Input : List = 2->3->4->5->4->3->2->1->NULL
Output : YES
Explanation : The difference between adjacent nodes in the list is 1. Hence the given list is a Jumper Sequence.
Input : List = 2->3->4->5->3->4->NULL
Output : NO
简单方法：
- 用一个临时指针遍历该列表。
- 将标志变量初始化为true，指示连续节点的绝对差为1。
- 开始遍历列表。
- 检查连续节点的绝对差是否为1。
- 如果是，则继续遍历，否则将标志变量更新为零并停止进一步遍历。
返回标志。
下面是上述方法的实现：
C++
// C++ program to check if absolute difference
// of consecutive nodes is 1 in Linked List
#include
using namespace std;
// A linked list node
struct Node {
int data;
struct Node* next;
};
// Utility function to create a new Node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
// Function to check if absolute difference
// of consecutive nodes is 1 in Linked List
bool isConsecutiveNodes(Node* head)
{
// Create a temporary pointer
// to traverse the list
Node* temp = head;
// Initialize a flag variable
int f = 1;
// Traverse through all the nodes
// in the list
while (temp) {
if (!temp->next)
break;
// count the number of jumper sequence
if (abs((temp->data) - (temp->next->data)) != 1) {
f = 0;
break;
}
temp = temp->next;
}
// return flag
return f;
}
// Driver code
int main()
{
// creating the linked list
Node* head = newNode(2);
head->next = newNode(3);
head->next->next = newNode(4);
head->next->next->next = newNode(5);
head->next->next->next->next = newNode(4);
head->next->next->next->next->next = newNode(3);
head->next->next->next->next->next->next = newNode(2);
head->next->next->next->next->next->next->next = newNode(1);
if (isConsecutiveNodes(head))
cout << "YES";
else
cout << "NO";
return 0;
} Java
// Java program to check if absolute difference
// of consecutive nodes is 1 in Linked List
class GFG
{
// A linked list node
static class Node
{
int data;
Node next;
};
// Utility function to create a new Node
static Node newNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.next = null;
return temp;
}
// Function to check if absolute difference
// of consecutive nodes is 1 in Linked List
static int isConsecutiveNodes(Node head)
{
// Create a temporary pointer
// to traverse the list
Node temp = head;
// Initialize a flag variable
int f = 1;
// Traverse through all the nodes
// in the list
while (temp != null)
{
if (temp.next == null)
break;
// count the number of jumper sequence
if (Math.abs((temp.data) - (temp.next.data)) != 1)
{
f = 0;
break;
}
temp = temp.next;
}
// return flag
return f;
}
// Driver code
public static void main(String[] args)
{
// creating the linked list
Node head = newNode(2);
head.next = newNode(3);
head.next.next = newNode(4);
head.next.next.next = newNode(5);
head.next.next.next.next = newNode(4);
head.next.next.next.next.next = newNode(3);
head.next.next.next.next.next.next = newNode(2);
head.next.next.next.next.next.next.next = newNode(1);
if (isConsecutiveNodes(head) == 1)
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code has been contributed by 29AjayKumarPython3
# Python3 program to check if absolute difference
# of consecutive nodes is 1 in Linked List
import math
# A linked list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Utility function to create a new Node
def newNode(data):
temp = Node(data)
temp.data = data
temp.next = None
return temp
# Function to check if absolute difference
# of consecutive nodes is 1 in Linked List
def isConsecutiveNodes(head):
# Create a temporary pointer
# to traverse the list
temp = head
# Initialize a flag variable
f = 1
# Traverse through all the nodes
# in the list
while (temp):
if (temp.next == None):
break
# count the number of jumper sequence
if (abs((temp.data) -
(temp.next.data)) != 1) :
f = 0
break
temp = temp.next
# return flag
return f
# Driver code
if __name__=='__main__':
# creating the linked list
head = newNode(2)
head.next = newNode(3)
head.next.next = newNode(4)
head.next.next.next = newNode(5)
head.next.next.next.next = newNode(4)
head.next.next.next.next.next = newNode(3)
head.next.next.next.next.next.next = newNode(2)
head.next.next.next.next.next.next.next = newNode(1)
if (isConsecutiveNodes(head)):
print("YES")
else:
print("NO")
# This code is contributed by SrathoreC#
// C# program to check if absolute difference
// of consecutive nodes is 1 in Linked List
using System;
public class GFG
{
// A linked list node
public class Node
{
public int data;
public Node next;
};
// Utility function to create a new Node
static Node newNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.next = null;
return temp;
}
// Function to check if absolute difference
// of consecutive nodes is 1 in Linked List
static int isConsecutiveNodes(Node head)
{
// Create a temporary pointer
// to traverse the list
Node temp = head;
// Initialize a flag variable
int f = 1;
// Traverse through all the nodes
// in the list
while (temp != null)
{
if (temp.next == null)
break;
// count the number of jumper sequence
if (Math.Abs((temp.data) - (temp.next.data)) != 1)
{
f = 0;
break;
}
temp = temp.next;
}
// return flag
return f;
}
// Driver code
public static void Main(String[] args)
{
// creating the linked list
Node head = newNode(2);
head.next = newNode(3);
head.next.next = newNode(4);
head.next.next.next = newNode(5);
head.next.next.next.next = newNode(4);
head.next.next.next.next.next = newNode(3);
head.next.next.next.next.next.next = newNode(2);
head.next.next.next.next.next.next.next = newNode(1);
if (isConsecutiveNodes(head) == 1)
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code contributed by Rajput-Ji输出：
YES
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