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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.
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Can you solve it without using extra space?
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查看是否有环，快慢两个指针一个移动一步，一个移动两步，是否相遇
查看环的起点，相遇后，将其中一个指针移到链表头，两个指针每次向后移动一步，相遇的点就是环的起点
证明可参考：http://www.cnblogs.com/wuyuegb2312/p/3183214.html
Linked List Cycle
bool hasCycle(ListNode *head) { if (head == nullptr || head->next == nullptr) return false; ListNode *slow = head, *fast = head; while (fast->next && fast->next->next) { fast = fast->next->next; slow = slow->next; if (fast == slow) return true; } return false; }
Linked List Cycle II
ListNode *detectCycle(ListNode *head) { if (head == nullptr || head->next == nullptr) return nullptr; ListNode *slow = head, *fast = head; while (fast->next && fast->next->next) { fast = fast->next->next; slow = slow->next; if (fast == slow) { slow = head; while (slow != fast) { slow = slow->next; fast = fast->next; } return slow; } } return nullptr; }
Linked List Cycle && Linked List Cycle II
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原文地址：http://www.cnblogs.com/sdlwlxf/p/5128498.html
