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?

/**
 * 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 ){  
            return null;  
        }  
    // 快指针fp和慢指针sp，  
    ListNode fp = head, sp = head;  
    while( fp != null && fp.next != null){  
        sp = sp.next;  
        fp = fp.next.next;  
        //此处应该用fp == sp ，而不能用fp.equals(sp) 因为链表为1 2 的时候容易  
        //抛出异常  
        if( fp == sp ){  //说明有环  
            break;  
        }  
    }  
    //System.out.println( fp.val + "   "+ sp.val );  
    if( fp == null || fp.next == null ){  
        return null;  
    }  
    //说明有环，求环的起始节点  
    sp = head;  
    while( fp != sp ){  
        sp = sp.next;  
        fp = fp.next;  
    }  
    return sp;  
    }  
}

/**
 * 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){
            return null;
        }
      
        ListNode fast=head, slow= head;
        while(fast!=null && fast.next!=null){
            fast=fast.next.next;
            slow =slow.next;
            if(fast==slow){
                break;
            }

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