剑指offer-链表中环的入口结点_牛客网

题目描述

给一个链表，若其中包含环，请找出该链表的环的入口结点，否则，输出null。

 

 

题目链接：

https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4?tpId=13&&tqId=11208&rp=1&ru=/activity/oj&qru=/ta/coding-interviews/question-ranking

 

 

分析：

空间换时间。

set记录之前出现过的节点。

 

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
import java.util.HashSet;
public class Solution {
    private HashSet<ListNode> set = new HashSet<>();
    public ListNode EntryNodeOfLoop(ListNode pHead)
    {
        ListNode cur = pHead;
        while(cur != null){
            if(set.contains(cur)){
                return cur;
            }
            set.add(cur);
            cur = cur.next;
        }
        return null;
    }
}

 

方法二：

快慢节点。详见：https://www.nowcoder.com/questionTerminal/253d2c59ec3e4bc68da16833f79a38e4 

 

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {
    public ListNode EntryNodeOfLoop(ListNode pHead)
    {
        if(pHead.next == null || pHead.next == null || pHead.next.next == null){
            return null;
        }
        //一开始需要先走一次？
        ListNode fast = pHead.next.next;
        ListNode slow = pHead.next;
        while(fast != slow){
            //快的走两步，慢的走一步
            if(fast.next != null && fast.next.next != null){
                fast = fast.next.next;
                slow = slow.next;
            }else{
                return null;
            }
        }
        fast = pHead;
        //x = (n - 2 * m )*c - a= (n - 2 *m -1 )*c + c - a
        //相遇节点即入口
        while(fast != slow){
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}