LeetCode //C - 160. Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

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.
   

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.

Constraints:

• The number of nodes of listA is in the m.
• The number of nodes of listB is in the n.
• $1<=m,n<=3\ast 10^4$
• $1<=Node.val<=10^5$
• 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.

From: LeetCode
Link: 160. Intersection of Two Linked Lists

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Solution:

Ideas：

1. Two Pointers: The algorithm initializes two pointers, one for each head of the linked lists (headA and headB).

2. Traversal: Both pointers advance through the lists, one node at a time.

3. Switching Lists: When a pointer reaches the end of a list, it continues from the beginning of the other list. This is a key step because it compensates for the difference in lengths between the two lists.

4. Equal Path Lengths: By switching lists, if there is an intersection, both pointers will eventually have traversed the same path length when they meet. This is because each pointer will have covered the length of its own list plus the length of the other list up to the intersection point.

5. Intersection Point: If the lists intersect, the pointers will meet at the intersection node. This happens because after switching, both pointers will have the same remaining distance to the intersection point. Since they move at the same pace, they must arrive there at the same time.

6. No Intersection: If the lists do not intersect, the pointers will end up being NULL at the same time, because they both will have traversed the full lengths of both lists.

7. Single Pass: Importantly, this approach only requires a single pass through each list, making it efficient. It doesn’t require knowledge of the lengths of the lists, nor does it require any additional data structures.

8. Efficiency: This method is very efficient in terms of time complexity, which is O(m+n), where m is the length of list A and n is the length of list B. The space complexity is O(1) since no extra space is used other than the pointers.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *ptrA = headA;
    struct ListNode *ptrB = headB;

    // Traverse both lists. Once reaching the end of a list, restart at the beginning of the other list.
    // If there is an intersection, the two pointers will eventually meet at the intersection node.
    // If there is no intersection, both pointers will eventually meet at the end of each other's list.
    while (ptrA != ptrB) {
        ptrA = ptrA ? ptrA->next : headB;
        ptrB = ptrB ? ptrB->next : headA;
    }

    // If the two pointers meet, ptrA (or ptrB) is the intersection node (or null if they do not intersect).
    return ptrA;
}
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