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.

题源：here；完整实现：here

思路：

通过两次遍历求出A和B两个链表的长度，然后在两者长度相等的情况下看其是否相等。

class Solution {
public:
	ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
		ListNode *pA, *pB;
		int lenA = 0, lenB = 0;
		pA = headA, pB = headB;
		while (pA) {
			lenA++;
			pA = pA->next;
		}
		while (pB) {
			lenB++;
			pB = pB->next;
		}
		pA = headA, pB = headB;
		while (pA && pB) {
			if (lenA > lenB) {
				lenA--;
				pA = pA->next;
				continue;
			}
			else if (lenA < lenB) {
				lenB--;
				pB = pB->next;
				continue;
			}
			else {
				if (pA == pB) return pA;
				else {
					pA = pA->next;
					pB = pB->next;
				}
			}
		}
		return NULL;
	}
};