算法打卡Day9_leetcode _142.环形链表 II

 //采用快慢双指针
    public ListNode detectCycle(ListNode head) {
        if (head == null) {
           return null;
        }
        ListNode fast = head;//快指针，每次走2步
        ListNode slow = head;//慢指针,每次走1步
        boolean ifCycle=false;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            //当快指针赶上慢指针，两指针有重叠时，说明是个环形链表
            if (slow == fast) {
                ifCycle = true;
                break;
            }
        }
        if (ifCycle){ //环存在
            slow =head ; //慢指针指向头节点
            while (slow!= fast){
                 fast =fast.next;
                 slow= slow.next;
            }
            return slow; //返回环开始的节点
        }
        return null;//环不存在
    }