Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
思路
这题是Linked List Cycle的进阶版
Given a linked list, determine if it has a cycle in it.
1 bool hasCycle(ListNode *head) { 2 if(head == NULL) return false; //带环链表还要考虑只有单个元素的情况 3 ListNode *faster = head, *slower = head; 4 while(faster->next != NULL && faster->next->next != NULL && slower->next != NULL){//直接判断faster->next->next != NULL会抛错 5 faster = faster->next->next; 6 slower = slower->next; 7 if(faster == slower) 8 return true; 9 } 10 return false; 11 }
faster和slower相遇之后必然在环上，让slower再走一圈计算环的长度len。另让两个指针p1,p2从head开始走，p1比p2先走len步，这样当p2走到环开始处时，正好p1与p2第一次相遇。
1 ListNode *detectCycle(ListNode *head) { 2 if(head == NULL) return NULL; 3 //带环链表要考虑只有单个元素的情况 4 ListNode *faster = head, *slower = head; 5 while(faster->next != NULL && faster->next->next != NULL && slower->next != NULL){//直接判断faster->next->next != NULL会抛错 6 faster = faster->next->next; 7 slower = slower->next; 8 if(faster == slower){ 9 ListNode *p1 = head;//另让两个指针p1,p2从head开始走 10 ListNode *p2 = head; 11 slower = slower->next;//让slower再走一圈计算环的长度len 12 p1 = p1->next;//设len是环的长度，p1比p2先走len步 13 if(faster == slower) return faster;//自环 14 while(faster != slower){ 15 slower = slower->next; 16 p1 = p1->next; 17 } 18 while(p1 != p2){//当p1与p2第一次相遇时，正好p2走到环开始处 19 p1 = p1->next; 20 p2 = p2->next; 21 } 22 return p2; 23 } 24 } 25 return NULL; 26 }