leetcode 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?

 

参考http://www.cnblogs.com/hiddenfox/p/3408931.html

 

方法：

第一次相遇时slow走过的距离：a+b，fast走过的距离：a+b+c+b。

因为fast的速度是slow的两倍，所以fast走的距离是slow的两倍，有 2(a+b) = a+b+c+b，可以得到a=c（这个结论很重要！）。

我们发现L=b+c=a+b，也就是说，从一开始到二者第一次相遇，循环的次数就等于环的长度。

我们已经得到了结论a=c，那么让两个指针分别从X和Z开始走，每次走一步，那么正好会在Y相遇！也就是环的第一个节点。

 

/**
 * 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(1 == 1){
            if(fast == null || fast.next ==null) return null;
            
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow){
                break;
            }
        }
        
        slow = head;
        while(fast != slow){
            fast = fast.next;
            slow = slow.next;
        }
        
        return slow;
        
    }
}

 

 

或者

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