## 1. 题目描述

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input 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 [5,0,1,8,4,5]. 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 = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input 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 [0,9,1,2,4]. 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: null
Input 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.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

## 2. 解题

### 方法1

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
return NULL;
}

// 计算两个链表长度的差值
int lenA = 0;
int lenB = 0;

while (ptrA) {
lenA++;
ptrA = ptrA->next;
}
while (ptrB) {
lenB++;
ptrB = ptrB->next;
}

// 让较长的链表先向后移动差值步
for (int i = 0; i < abs(lenA - lenB); i++) {
if ((lenA - lenB) < 0) {
ptrB = ptrB->next;
}
else {
ptrA = ptrA->next;
}
}

while (1) {
if (ptrA == ptrB) {
return ptrA;
}
else if (ptrA == NULL || ptrB == NULL) {
return NULL;
}
else {
ptrA = ptrA->next;
ptrB = ptrB->next;
}
}
}

### 方法2

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
}