标签：border blog round 环入口 ret arc def ack tps
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.
想法：（1）首先的判断链表中是否有环，若有环进行(2),没有环就返回NULL
(2)参考https://www.cnblogs.com/jack204/archive/2011/09/14/2175559.html的分析，
从链表头到环入口点等于(n-1)循环内环+相遇点到环入口点。于是可以从链表头和相遇点分别设一个 指针，每次各走一步，两个指针必定相遇，且相遇第一点为环入口点。此时返回其中一个指针即可。
/** * 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) { if(NULL == head || head->next == NULL) return NULL; bool cycle = false; struct ListNode* fast = head; struct ListNode* slow = head; while(fast && fast->next){ slow = slow->next; fast = fast->next->next; if(slow == fast){ cycle = true; break; } } if(cycle){ slow = head; while(slow != fast){ slow = slow->next; fast = fast->next; } return slow; } return NULL; } };
leetcode-142 Linked List Cycle II
标签：border blog round 环入口 ret arc def ack tps
原文地址：https://www.cnblogs.com/tingweichen/p/9882361.html
