141.Linked List Cycle 快慢指針
問題描述:
Given a linked list, determine if it has a cycle in it.
我的思路:
用一個指針指向鏈表頭部,另一個指針在鏈表上移動。若兩個指針指向地址相等,則存在環。
然後我沒過OJ:)
因為循環鏈表的尾部並不一定鏈接頭部,可能鏈接任意節點,此時會陷入死循環。
所以應用快慢指針。
兩個指針的步長不同,若存在環,必然會相遇。
代碼實現(c++):
class Solution { public: bool hasCycle(ListNode *head) { ListNode *p1 = head, *p2 = head;while (p2 && p2->next) { p1 = p1->next; p2 = p2->next->next; if (p1 == p2) return true; } return false; } };
141.Linked List Cycle 快慢指針
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