153 lines
4.7 KiB
Markdown
153 lines
4.7 KiB
Markdown
- # Leetcode Intersection-of-Two-Linked-Lists
|
|
|
|
2022-09-08 16:13
|
|
|
|
> ##### Algorithms:
|
|
>
|
|
> #algorithm #two_pointers
|
|
>
|
|
> ##### Data structures:
|
|
>
|
|
> #DS #linked_list #set
|
|
>
|
|
> ##### Difficulty:
|
|
>
|
|
> #coding_problems #difficulty_easy
|
|
>
|
|
> ##### Additional tags:
|
|
>
|
|
> #leetcode
|
|
>
|
|
> ##### Revisions:
|
|
>
|
|
> N/A
|
|
|
|
##### Links:
|
|
|
|
- [Link to problem](https://leetcode.com/problems/intersection-of-two-linked-lists/)
|
|
|
|
***
|
|
|
|
### Problem
|
|
|
|
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**.
|
|
|
|
#### 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.
|
|
|
|
#### 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 * 104`
|
|
- `1 <= Node.val <= 105`
|
|
- `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.
|
|
|
|
### Thoughts
|
|
|
|
> [!summary]
|
|
> This can be solved using #two_pointers or #set
|
|
|
|
#### Set
|
|
|
|
Record the address of visited node iterating over `listA`,
|
|
and verify if the address has been visited when iterating
|
|
over `listB`
|
|
|
|
#### Two pointers
|
|
|
|
#TODO Use two pointers method to solve
|
|
|
|
### Solution
|
|
|
|
#### Set
|
|
|
|
```cpp
|
|
/**
|
|
* Definition for singly-linked list.
|
|
* struct ListNode {
|
|
* int val;
|
|
* ListNode *next;
|
|
* ListNode(int x) : val(x), next(NULL) {}
|
|
* };
|
|
*/
|
|
class Solution {
|
|
public:
|
|
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
|
|
// O(M + N), using set.
|
|
unordered_set<ListNode *> used;
|
|
|
|
ListNode *ptrA = headA;
|
|
|
|
while (ptrA) {
|
|
used.insert(ptrA);
|
|
ptrA = ptrA->next;
|
|
}
|
|
|
|
ListNode *ptrB = headB;
|
|
|
|
while (ptrB) {
|
|
if (used.find(ptrB) != used.end()) {
|
|
// intersect
|
|
return ptrB;
|
|
}
|
|
|
|
ptrB = ptrB->next;
|
|
}
|
|
|
|
return nullptr;
|
|
}
|
|
};
|
|
```
|