<LeetCode OJ> 141 / 142 Linked List Cycle(I / II)
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
分析:
假设有环?遍历链表将无法走完,假设无环终会走到尾为NULL的位置
让一个指针每次走一个,一个指针每次走两个位置。
假设当中一个为NULL则无环。
假设相遇(必会相遇)了则有环。
time,o(n),space,o(1)
/** * 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==NULL) return false; ListNode *showNode=head; ListNode *fastNode=head; while(true) { if(showNode->next!=NULL) showNode=showNode->next; else return false; if(fastNode->next!=NULL && fastNode->next->next!=NULL) fastNode=fastNode->next->next; else return false; if(showNode==fastNode) return true; } return false; } };
别人的简洁算法:一样的思路
class Solution { public: bool hasCycle(ListNode *head) { ListNode *slow=head,*fast=head; while(slow&&fast&&fast->next){ slow=slow->next; //跑的慢 fast=fast->next->next; //跑的快 if(slow==fast) return true; //相遇则有环 } return false; } };
Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
分析:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *detectCycle(ListNode *head) { ListNode *slow=head,*fast=head; while(slow && fast && fast->next){ slow=slow->next; //跑的慢 fast=fast->next->next; //跑的快 if(slow==fast) break; //相遇则有环 } if(fast==NULL || fast->next==NULL) return NULL; fast=head; while(slow != fast){ slow=slow->next; //一样的速度跑 fast=fast->next; } return slow; } };
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原文地址:http://blog.csdn.net/ebowtang/article/details/50507131
原作者博客:http://blog.csdn.net/ebowtang
本博客LeetCode题解索引:http://blog.csdn.net/ebowtang/article/details/50668895