Problem 160: Intersection Of Two Linked Lists

Question

LeetCode Link | 160. Intersection Of Two Linked Lists | Easy

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

For example, the following two linked lists begin to intersect at node c1:

The test cases are generated such that there are no cycles anywhere in the entire linked structure.

Note that the linked lists must retain their original structure after the function returns.

Custom Judge:

The inputs to the judge are given as follows (your program is not given these inputs):

• intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersected node.
• listA - The first linked list.
• listB - The second linked list.
• skipA - The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
• skipB - The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.

The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted.

Constraints

• The number of nodes of listA is in the m.
• The number of nodes of listB is in the n.
• 1 <= m, n <= 3 * 10^4
• 1 <= Node.val <= 10^5
• 0 <= skipA < m
• 0 <= skipB < n
• intersectVal is 0 if listA and listB do not intersect.
• intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.

Examples

Example 1

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at ‘8’
Explanation: The intersected node’s value is 8 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

• Note that the intersected node’s value is not 1 because the nodes with value 1 in A and B (2nd node in A and 3rd node in B) are different node references. In other words, they point to two different locations in memory, while the nodes with value 8 in A and B (3rd node in A and 4th node in B) point to the same location in memory.

Example 2

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at ‘2’
Explanation: The intersected node’s value is 2 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: No intersection
Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Solution Approach

Method: Manual / Simulate

Constrains

• None

This method does not have any limitations.

Concept Explanation

The simplest way is iterate one list and store all the nodes in stack and iterate the second list to check if any node appears in the stack. However, this cost more space used as O(n), and it can be reduced to O(1), here is how to do it:

It’s not hard to find that no matter what the lists look like, they all have same ending if they are intersected since once intersect all the node are the same. Thus, we could just align these two lists at the end, and check node if intersected index by index as the figure shown below:

It should be soon to find the intersection as it check to the next node, but how do we know where to place these two pointers?

The answer is to do this by hand, we have to firstly get two list’s length (which is [ListA: 5, ListB: 3] in this example) and then find the gap between them (which is 2), and finally, you know curA has to be move 2 nodes as the start to match the curB pointer in listB.

Now it’s simple, just check if curA is equal to curB and iterate to the next node before they reach the end of the list. If no node find is equal then return null.

Code

• The number of nodes of listA is in the m.
• The number of nodes of listB is in the n.
• Time complexity: O(m+n)
• Space complexity: O(1)

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
    if (!headA || !headB) return null;
    let sizeA: number = 0, sizeB: number = 0;
    let curA: ListNode | null = headA, curB: ListNode | null = headB;
    // Get Length of each list
    while (curA) {
        sizeA++;
        curA = curA.next;
    }
    while (curB) {
        sizeB++;
        curB = curB.next;
    }

    // Reset iterator node
    curA = headA;
    curB = headB;

    // Swap list, make sure A is the long list
    if (sizeA < sizeB) {
        [sizeA, sizeB] = [sizeB, sizeA];
        [curA, curB] = [curB, curA];
    }

    // Make two list align at the end
    let gap = sizeA - sizeB;
    while (gap-- && curA) {
        curA = curA.next;
    }

    // Find intersection
    while (curA && curB) {
        if (curA === curB) return curA;
        curA = curA.next;
        curB = curB.next;
    }
    return null;
};

Conclusion

A easy practice for linked list, just know that to check if node the same are checking the address of the node, not the value of the node.