# 160. Intersection of Two Linked Lists

### Leetcode

<https://leetcode.com/problems/intersection-of-two-linked-lists/>

### 題目

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`:

![](https://assets.leetcode.com/uploads/2021/03/05/160_statement.png)

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**.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_1_1.png)

```
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.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_2.png)

```
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:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_3.png)

```
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.
```

### 解答 <a href="#ti-jie" id="ti-jie"></a>

* **方法一**

hashmap 法

```
var getIntersectionNode = function(headA, headB) {
    const map = new Map();
    
    while(headA || headB) {
        if(headA === headB) return headA;
        if(map.has(headA)) return headA;
        if(map.has(headB)) return headB;
            
        headA && map.set(headA, true);
        headB && map.set(headB, true);
        headA = headA?.next;
        headB = headB?.next;
    }
    
    return null;
};
```

* **方法二**

two pointer

用 `pA`、`pB` 遍歷 linked list，pA 遍歷到底時指向 headB，pB 遍歷到底時指向 pA，如此在遍歷第二輪時兩個 pointer 就可以指向同一個順序的元素，此時就只要比較 pA, pB 是否相等即可

```
var getIntersectionNode = function(headA, headB) {
    let pA = headA;
    let pB = headB;
    
    while(pA !== pB) {
        pA = pA === null ? headB : pA.next;
        pB = pB === null ? headA : pB.next;
    }
    
    // 有可能此時 pA 是某個 intersection 或者 第二輪遍歷完後 pA === null 
    return pA;
};
```

Runtime: 178 ms, faster than 14.66% of JavaScript online submissions for Intersection of Two Linked Lists.

Memory Usage: 45.1 MB, less than 98.88% of JavaScript online submissions for Intersection of Two Linked Lists.
