leetcode 142. Linked List Cycle II ----- java

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?

141题的延伸,求出循环点。

可以用数学方法证明出slow与find相遇的位置一定是所求的点。

/**
 * 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;
        ListNode slow = head;

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

        return null;

    }
}
时间: 2024-10-13 11:48:40

leetcode 142. Linked List Cycle II ----- java的相关文章

Java for LeetCode 142 Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 解题思路,本题和上题十分类似,但是需要观察出一个规律,参考LeetCode:Linked List Cycle II JAVA实现如下: public ListNode detectCycle(Li

Leetcode 142 Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 同Leetcode 141 Linked List Cycle 性质:distance from head to 环开始点 == distance from 双指针相遇的点 to 环开始点 证明: 指

[LeetCode 题解]: Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 题意: 给定一个链表,找到环起始的位置.如果环不存在,返回NULL. 分析: (1)首先要判断该链表是否有环.如果没有环,那么返回NULL. (2)其次,当已知环存在后,寻找环起始的位置. 思路: (

【Leetcode】Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 思路:由[Leetcode]Linked List Cycle可知,利用一快一慢两个指针能够判断出链表是否存在环路.假设两个指针相遇之前slow走了s步,则fast走了2s步,并且fast已经在长度

LeetCode OJ - Linked List Cycle II

题目: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 解题思路: 设置两个指针slow和fast,从head开始,slow一次一步,fast一次两步,如果fast能再次追上slow则有圈. 设slow走了n步,则fast走了2*n步,设圈长度m

141. Linked List Cycle && 142. Linked List Cycle II

141. Linked List Cycle 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 { pu

刷题142. Linked List Cycle II

一.题目说明 题目142. Linked List Cycle II,判断一个链表是否有环,如果有返回环的第一个元素,否则返回NULL. 这个题目是141. Linked List Cycle的升级版本,难度是Medium! 二.我的解答 最直观的解答就是用一个unordered_map<ListNode*,int> dp来统计节点出现的次数,如果出现2,则这个就是第一个节点. class Solution{ public: ListNode* detectCycle(ListNode* he

leetcode:142. Linked List Cycle II(Java)解答

转载请注明出处:z_zhaojun的博客 原文地址:http://blog.csdn.net/u012975705/article/details/50412899 题目地址:https://leetcode.com/problems/linked-list-cycle-ii/ Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return

[LeetCode][JavaScript]Linked List Cycle II

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? https://leetcode.com/problems/linked-list-