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题目:
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
代码:
用hashmap版
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { std::map<ListNode *, bool> ifNodeOccured; ListNode *p = head; while ( p ) { if ( ifNodeOccured.find(p) != ifNodeOccured.end() ) return true; ifNodeOccured.insert(std::pair<ListNode *, bool>(p,true)); p = p->next; } return false; } };
不用hashmap,改用双指针技巧版
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { if (!head) return false; ListNode *p1 = head; ListNode *p2 = head; while ( p2->next && p2->next->next ) { p2 = p2->next->next; p1 = p1->next; if (p2==p1) return true; } return false; } };
Tips:
这两种解法都是基本的技巧。第二种双指针的技巧效率更高,且空间复杂度更低,似乎得到了更多的推崇;但个人总觉得hashmap的方法不错,而且不强依赖于技巧。
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原文地址:http://www.cnblogs.com/xbf9xbf/p/4470102.html