Linked lists are a fundamental data structure in computer science, widely used in various applications such as databases, programming languages, and operating systems. One of the most common operations that can be performed on linked lists is finding the intersection point of two linked lists. In this article, we will discuss the concept of linked lists, their intersection point, and how to find the intersection point of two linked lists in C++.
A linked list is a sequence of data elements, each containing a reference to the next element in the list. The first element in the list is called the head, and the last element is called the tail. Linked lists are often used because they are dynamic in nature, meaning that they can grow or shrink in size as needed. They are also useful for storing large amounts of data because they do not have a fixed size like arrays.
The intersection point of two linked lists is the point where the two lists merge. In other words, it is the point where two lists share the same node. For example, consider the following two linked lists:
List 1: 1 -> 2 -> 3 -> 4 -> 5 -> 6 List 2: 7 -> 8 -> 9 -> 4 -> 5 -> 6
In this example, the intersection point of the two lists is the node containing the value 4.
To find the intersection point of two linked lists in C++, we can use a simple algorithm known as the "two pointer" method. This algorithm involves using two pointers, one for each list, and moving them through the lists until they reach the intersection point. The basic idea behind this algorithm is that if two linked lists intersect, they will have the same tail node. Therefore, we can simply compare the tail nodes of the two lists, and if they are the same, we know that the lists intersect.
Here is an example of how to find the intersection point of two linked lists in C++ using the two pointer method:
#include <iostream> using namespace std; // Definition for singly-linked list struct ListNode { int val;
ListNode *next; ListNode(int x) : val(x), next(NULL) {}
}; ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) { // Initialize pointers ListNode* pointerA = headA;
ListNode* pointerB = headB; // Traverse both lists until we reach the end while (pointerA != pointerB) { pointerA = pointerA ? pointerA->next : headB; pointerB = pointerB ? pointerB->next : headA; }
return pointerA; }
In this example, we first initialize two pointers, pointerA and pointerB, to the heads of the two linked lists. We then traverse both lists using a while loop until we reach the end. Inside the loop, we check if the current pointer is at the end of its list. If it is, we set it to the head of the other list. This allows us to traverse both lists in parallel, and if the lists intersect, the pointers will eventually meet at the intersection point.
One important thing to note is that the time complexity of this algorithm is O(n), where n is the total number of nodes in both lists. This is because we are traversing both lists once, and the total number of nodes in both lists is equal to the sum of the number of nodes in each list.
Another important thing to consider is that this algorithm assumes that the
