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 = 3Output: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:**

Input:intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1Output: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 = 2Output:No intersectionExplanation: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`

. `0 <= 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.

**Follow up:** Could you write a solution that runs in `O(n)`

time and use only `O(1)`

memory?

**Solution 1: Two Passes by swapping heads**

Time complexity: O(n)

Space complexity: O(1)

## C++

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// Author: Huahua class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { if (!headA || !headB) return nullptr; ListNode* a = headA; ListNode* b = headB; while (a != b) { a = a ? a->next : headB; b = b ? b->next : headA; } return a; } }; |