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?

思路

首先证明链表有环，再让一个指针从头开始，另一个指针从相遇的位置，每次都只走一步，相遇的地方就是链表的环开始的地方。

代码

public ListNode detectCycle(ListNode head) {    if (head == null) return null;    boolean flag = false;    ListNode fast = head, slow = head;    while (fast != null && fast.next != null) {        fast = fast.next.next;        slow = slow.next;        if (fast == slow) {            flag = true;            break;        }    }    if (!flag) return null;    slow = head;    while (true) {        //先判断，避免只有两个节点的环的情况        if (slow == fast) return slow;        slow = slow.next;        fast = fast.next;    }}