Linked List Cycle

Given a linked list, determine if it has a cycle in it.

链表判断是否为环。用快慢指针，快的一次走两步，慢的一次走一步。如果快的能追上慢的那说明有环。

public boolean hasCycle(ListNode head) {
    ListNode fast = head;
    ListNode slow = head;

    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
        if (fast == slow) {
            return true;
        }
    }
    return false;
}

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.

接上题，如果有环就找到环的起点。

数学证明略。还是快慢指针，当它们第一次相遇的时候把慢的那根指针重新指向链表头。它们再次相遇的那个点就是环的起点。

public ListNode detectCycle(ListNode head) {
    ListNode fast = head;
    ListNode slow = head;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
        if (fast == slow) {
            slow = head;
            while (slow != fast) {
                fast = fast.next;
                slow = slow.next;
            }
            return slow;
        }
    }
    return null;
}