2019.7.23

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.（链表中有没有环的问题）

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head == null || head.next == null || head.next.next == null){
            return null;
        }
        ListNode slow = head.next;
        ListNode fast = head.next.next;
        while(fast != slow){
                if(fast.next == null || fast.next.next == null){
                    return null;
                }
                fast = fast.next.next;
                slow = slow.next;
        }
        fast = head;
        while(fast != slow){
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

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

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null || head.next == null || head.next.next == null){
            return false;
        }
        ListNode fast = head.next.next;
        ListNode slow = head.next;
        while(fast != slow){
            if(slow.next == null || fast.next == null || fast.next.next == null){
                return false;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        fast = head;
        while(fast != slow){
            fast = fast.next;
            slow = slow.next;
        }
        return true;
    }
}