赶新鲜，和solution的解法不太一样，解三元一次方程组。。。
简而言之，list A的长度是x+z, listB的长度是y+z. z为两个list共同部分的长度。把A reverse一下，可以得到x+y的值，然后算一下就都出来了。
ListNode* reverseList(ListNode* head){if (!head){return NULL;}if (!head->next){return head;}ListNode* pre = NULL;ListNode* cur = head;while (cur->next){ListNode* tmp = cur->next;cur->next = pre;pre = cur;cur = tmp;}cur->next = pre;return cur;
}ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {if (!headA || !headB){return NULL;}ListNode* pa = headA;int countX = 1;while (pa->next){pa = pa->next;countX++;}ListNode* pb = headB;int countY = 1;while (pb->next){pb = pb->next;countY++;}if (pa != pb){return NULL;}ListNode* tmp = reverseList(headA);pb = headB;int countZ = 0;while (pb->next){pb = pb->next;countZ++;}reverseList(tmp);int a = (countX - countY + countZ) / 2;pa = headA;while (a--){pa = pa->next;}return pa;
}