leetcode 141、Linked list cycle

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leetcode 141、Linked list cycle

一种方法是用set存储出现过的指针,重复出现时就是有环: class Solution { public: bool hasCycle(ListNode *head) { set<ListNode*> st; ListNode *p = head; while(p) { if (st.find(p) != st.end()) return true; st.insert(p); p = p->next; } return false; } }; 另一种方法就是快慢指针,快指针一次走两步,…

[LeetCode] 141&142 Linked List Cycle I & II

Problem: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? https://oj.leetcode.com/problems/linked-list-cycle/ Problem II: Given a linked list, return the node where the cycle begins. If the…

leetcode 141 142. Linked List Cycle

题目描述: 不用辅助空间判断,链表中是否有环 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /*一个指针走的快一个指针走得慢*/ class Solution { public: bool hasCycle(ListNode *head) { if(head ==…

LeetCode解题报告:Linked List Cycle && Linked List Cycle II

LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Linked List Cycle II Given a linked list, return the node w…

LeetCode解题报告—— Linked List Cycle II & Reverse Words in a String & Fraction to Recurring Decimal

1. Linked List Cycle II 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. 思路:想法是利用两指针,一个每次移动一步,另一个每次移动两步,如果存在环则这两个指针一定会相遇(这里可以在纸上画一下,因为后一个指针移动比前一个指针快,当后一个指针在环中来到前一个指针的…

【LeetCode】142. Linked List Cycle II

Difficulty:medium More:[目录]LeetCode Java实现 Description 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? Intuiti…

【LeetCode】142. Linked List Cycle II (2 solutions)

Linked List Cycle II 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? 解法一: 使用unordered_map记录当前节点是否被访问过,如访问过返回该节点,如到达尾部说明无环. /** * Definition for sing…