[LeetCode]142. Linked List Cycle II

142. 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.

Follow up:
Can you solve it without using extra space?

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/** * 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) {        // 1.找到环中的任意一个节点        ListNode* oneNode = oneNodeInCycle(head);        if(oneNode == NULL)            return NULL;        // 2.计算环的长度        int len = 1;        ListNode* temp = oneNode;        while(oneNode->next != temp){            ++len;            oneNode = oneNode->next;        }        // 3.让fast指针先行len步        ListNode* fast = head;        ListNode* slow = head;        while(len--){            fast = fast->next;        }        // 4.fast和slow指针下次相遇的节点就是入口节点        while(fast != slow){            fast = fast->next;            slow = slow->next;        }        return fast;    }    // 如果有环，返回环中一个节点    // 如果没有环，返回NULL    ListNode* oneNodeInCycle(ListNode* head){        ListNode* fast = head;        ListNode* slow = head;        while(fast && fast->next){            fast = fast->next->next;            slow = slow->next;            if(fast == slow)                return fast;        }        return NULL;    }};