Intersection of Two Linked Lists

class Solution {
public:
    ListNode* findIntersection(ListNode* startA, ListNode* startB) {
        ListNode* currentA = startA;
        ListNode* currentB = startB;
        while (currentA != currentB) {
            currentA = currentA == nullptr ? startB : currentA->next;
            currentB = currentB == nullptr ? startA : currentB->next;
        }
        return currentA;
    }
};

This solution resolves the problem of finding the intersection node of two singly linked lists. The code, written in C++, defines a method findIntersection within the class Solution, which returns the intersected node if an intersection exists, or nullptr otherwise.

• The function uses two pointers, currentA and currentB, initialized at the heads of the two lists, startA and startB.
• Both pointers traverse through their respective lists, and once they reach the end, they continue from the start of the other list.
• This step ensures that both pointers cover the same distance, taking into account the difference in lengths of the two lists.
• If the lists intersect, the pointers will meet at the intersection node after a specific number of steps; otherwise, they will both reach the end (nullptr), indicating no intersection.

This approach leverages the cyclical traversal by the two pointers to equate their paths, allowing them to find the intersection in linear time and constant space.

public class Solution {
    public ListNode findIntersection(ListNode firstHead, ListNode secondHead) {
        ListNode firstPointer = firstHead;
        ListNode secondPointer = secondHead;
        while (firstPointer != secondPointer) {
            firstPointer = firstPointer == null ? secondHead : firstPointer.next;
            secondPointer = secondPointer == null ? firstHead : secondPointer.next;
        }
        return firstPointer;
    }
}

The provided Java solution identifies the intersection point of two singly linked lists. The technique employed ensures that no additional memory is used, maintaining time complexity at O(m+n), where m and n are the lengths of the two linked lists.

In the solution:

• Define a method findIntersection that takes the heads of two linked lists (firstHead and secondHead) as parameters.
• Initialize two pointers, firstPointer and secondPointer, to start at the heads of the two lists respectively.
• Utilize a while loop to traverse the lists. The loop continues until the two pointers meet, indicating the intersection point, or if there's no intersection, until both pointers reach the end of the lists (both are null).
• Inside the loop, move each pointer to the next node in its respective list. If a pointer reaches the end of its list, redirect it to the head of the other list.
• The looping mechanism inherently handles the difference in lengths between the two lists by shifting pointers to the start of the opposite list. This guarantee, after traversing at most 2*(m+n) steps, either an intersection is found, or both pointers simultaneously reach the end of their paths.

At the conclusion of the loop, firstPointer will either point to the intersection node, or be null if the lists do not intersect. This efficient method effectively addresses the problem without modifying the original lists or incurring additional space complexity.

struct ListNode *findIntersection(struct ListNode *list1, struct ListNode *list2) {
    struct ListNode *pointer1 = list1;
    struct ListNode *pointer2 = list2;
    while (pointer1 != pointer2) {
        pointer1 = pointer1 == NULL ? list2 : pointer1->next;
        pointer2 = pointer2 == NULL ? list1 : pointer2->next;
    }
    return pointer1;
}

The provided C solution effectively identifies the intersection node of two singly linked lists. The core approach utilizes two pointers, pointer1 and pointer2, each initially set to the heads of list1 and list2 respectively.

• Each pointer progresses through its respective list.

• Once a pointer reaches the end of its list, it switches to the beginning of the other list.

• This crisscrossing continues until both pointers meet at the intersection node or both reach the end (NULL), indicating no intersection.

The main strength of this approach lies in its guarantee to find the intersection in linear time, O(n + m), where n and m are the lengths of the lists. This eliminates the need for additional data structures, making the implementation both time-efficient and space-efficient without additional memory overhead.

This algorithmic solution leverages the concept of two-pointer technique to solve the problem effectively with a simple but clever approach. The conditional operator is used to loop back the pointers to the start of the opposite list once the end of one list is reached, ensuring that any difference in the lengths of the lists is accounted for inherently. This method ensures that if an intersection exists, the pointers will align at the intersection node simultaneously.

var findIntersection = function (list1, list2) {
    var current1 = list1;
    var current2 = list2;
    while (current1 !== current2) {
        current1 = current1 === null ? list2 : current1.next;
        current2 = current2 === null ? list1 : current2.next;
    }
    return current1;
};

In the provided JavaScript solution, the problem of finding the intersection point of two singly linked lists is tackled effectively. Here’s a streamlined explanation of how the given function, findIntersection, works:

• Initialize two pointers, current1 and current2, that traverse through list1 and list2 respectively.
• Use a while loop that continues until these two pointers meet, which will happen at the intersection node or when both are null (i.e., no intersection).
• Inside the loop, advance each pointer to the next node in its respective list. If a pointer reaches the end of its list (null), it starts over at the beginning of the other list.
• When current1 and current2 finally meet, the loop breaks and the method returns current1 (or current2, since both are the same at this point) as the intersecting node. If no intersection exists, it returns null.

This solution leverages the property that by switching the list heads, both pointers will have traversed the same number of nodes (the length of both lists combined) when they meet, thus assuring they meet at the intersection point or conclude with no intersection.

This approach is both space-efficient, requiring no additional data structures, and time-efficient, with a linear time complexity relative to the sum of the lengths of the two lists. Adjustments to the runtime environment or the data structure may require minor adaptations of the function for optimal performance.

class Solution:
    def findIntersection(self, list1: ListNode, list2: ListNode) -> ListNode:
        pointer1 = list1
        pointer2 = list2

        while pointer1 != pointer2:
            pointer1 = list2 if pointer1 is None else pointer1.next
            pointer2 = list1 if pointer2 is None else pointer2.next

        return pointer1

The provided Python code defines a function to find the intersection point of two singly linked lists. The function, findIntersection, utilizes two pointers, pointer1 and pointer2, each starting at the head of the respective linked lists, list1 and list2.

• At each step, both pointers move one node forward.
• If a pointer reaches the end of its list, it then moves to the beginning of the other list.
• This continues until the pointers meet, which will be the intersection node or None if there is no intersection.

This approach ensures that both pointers traverse exactly len(list1) + len(list2) nodes. By switching lists, any difference in the length of the lists is counterbalanced, and if the lists intersect, the pointers will eventually meet at the intersection node. The solution returns the node at which both pointers coincide, thus indicating the intersection point. If no intersection exists, the function returns None.