题目分析：两个链表如果相交且不存在环，那么这两个链表从相交节点往后的节点都相同，所以，遍历一个链表，在遍历时不断遍历另一个链表，只要相等就可以返回了

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {struct ListNode*point1=headA;struct ListNode*point2=headB;if(headA==NULL||headB==NULL){return NULL;}while(point2!=NULL){   point1=headA;while(point1!=NULL){if(point1==point2){return point2;}point1=point1->next;}point2=point2->next;}return NULL;
}