[LeetCode] Linked List Cycle II, Solution
2013-11-24 06:11
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Given a linked list, return the node where the cycle begins. If there is no cycle, return
Can you solve it without using extra space? [Thoughts] 首先,比较直观的是,先使用Linked List Cycle I的办法,判断是否有cycle。如果有,则从头遍历节点,对于每一个节点,查询是否在环里面,是个O(n^2)的法子。但是仔细想一想,发现这是个数学题。 如下图,假设linked list有环,环长Y,环以外的长度是X。
现在有两个指针,第一个指针,每走一次走一步,第二个指针每走一次走两步,如果他们走了t次之后相遇在K点
那么 指针一 走的路是 t = X + nY + K ①
指针二 走的路是 2t = X + mY+ K ② m,n为未知数
把等式一代入到等式二中, 有
2X + 2nY + 2K = X + mY + K
=> X+K = (m-2n)Y ③
这就清晰了,X和K的关系是基于Y互补的。等于说,两个指针相遇以后,再往下走X步就回到Cycle的起点了。这就可以有O(n)的实现了。
[code] ListNode *detectCycle(ListNode *head) { ListNode * first = head; ListNode * second = head; while(first != NULL && second != NULL) { first = first->next; second = second->next; if(second != NULL) second = second->next; if(first == second) break; } if(second == NULL) return NULL; // 一起往下走X步,就找到节点了。 first = head; while(first!=second) { first = first->next; second = second->next; } return second; }
null. Follow up:
Can you solve it without using extra space? [Thoughts] 首先,比较直观的是,先使用Linked List Cycle I的办法,判断是否有cycle。如果有,则从头遍历节点,对于每一个节点,查询是否在环里面,是个O(n^2)的法子。但是仔细想一想,发现这是个数学题。 如下图,假设linked list有环,环长Y,环以外的长度是X。
现在有两个指针,第一个指针,每走一次走一步,第二个指针每走一次走两步,如果他们走了t次之后相遇在K点
那么 指针一 走的路是 t = X + nY + K ①
指针二 走的路是 2t = X + mY+ K ② m,n为未知数
把等式一代入到等式二中, 有
2X + 2nY + 2K = X + mY + K
=> X+K = (m-2n)Y ③
这就清晰了,X和K的关系是基于Y互补的。等于说,两个指针相遇以后,再往下走X步就回到Cycle的起点了。这就可以有O(n)的实现了。
[code] ListNode *detectCycle(ListNode *head) { ListNode * first = head; ListNode * second = head; while(first != NULL && second != NULL) { first = first->next; second = second->next; if(second != NULL) second = second->next; if(first == second) break; } if(second == NULL) return NULL; // 一起往下走X步,就找到节点了。 first = head; while(first!=second) { first = first->next; second = second->next; } return second; }
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