【Leetcode】【数据结构】【C语言】判断两个链表是否相交并返回交点地址

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {struct ListNode *tailA=headA;struct ListNode *tailB=headB;int count1=0;int count2=0;//分别找尾节点，并顺便统计节点数量：while(tailA){tailA=tailA->next;count1++;}while(tailB){tailB=tailB->next;count2++;}if(tailA!=tailB)//如果尾节点不相同，则一定不相交return NULL;int tmp=abs(count1-count2);//得到两个数量差值的绝对值struct ListNode*longList=headA;struct ListNode*shortList=headB;//找出长的链表：if(count2>count1){longList=headB;shortList=headA;}//先让长的链表走差值步，再和短链表齐头并进while(tmp){longList=longList->next;tmp--;}//两个链表齐头并进：while(longList!=shortList){shortList=shortList->next;longList=longList->next;}//两个链表相遇的地方就是节点return longList;
}