Linked List Cycle && 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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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;
}