Linked List Cycle II 题解 题目来源:https://leetcode.com/problems/linked-list-cycle-ii/description/ 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: C…

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? 解题思路,本题和上题十分类似,但是需要观察出一个规律,参考LeetCode:Linked List Cycle II JAVA实现如下: public ListNode detectCycle(Li…

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? 思路:由[Leetcode]Linked List Cycle可知.利用一快一慢两个指针可以推断出链表是否存在环路. 如果两个指针相遇之前slow走了s步,则fast走了2s步.而且fast已经在长…

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? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

Linked List Cycle 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/linked-list-cycle/description/ Description Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Solution class Solution { publ…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to.…

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? 思路 这题是Linked List Cycle的进阶版 Given a linked list, determine if it has a cycle in it. bool hasCycle(Li…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to.…

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? 参考http://www.cnblogs.com/hiddenfox/p/3408931.html 方法: 第一次相遇时slo…

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? 思路: 做过,就当复习了. 先用快慢指针判断相交,关键是环开始点的获取. 用上图说明一下,设非环的部分长度为a(包括环的入口点), 环的长度为b(包括环的入口点). 快慢指针相交的位置为绿色的点,距离…