Lintcode阶梯训练~算法

带环链表2

2017-06-28  本文已影响21人  lyoungzzz

描述

给定一个链表,如果链表中存在环,则返回到链表中环的起始节点的值,如果没有环,返回 null。

样例

给出 -21->10->4->5, tail connects to node index 1,返回 10

代码实现

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: The node where the cycle begins. 
     *           if there is no cycle, return null
     */
    public ListNode detectCycle(ListNode head) {  
        if (head == null || head.next == null) {
            return null;
        }
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != slow) {
            if (fast == null || fast.next == null) {
                return null;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        /**while (head != fast.next) {
            head = head.next;
            slow = fast.next;
        }**/
//用慢指针的next与head比较
        while (head != slow.next) {
            head = head.next;
            slow = slow.next;
        }
        return head;
    }
}

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