/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *p1, *p2; //p1和p2从链表的第一个节点出发,p1每次移动一个节点,p2每次移动两个节点,若两个指针重逢,则说明有环存在,否则若p2遇到NULL指针,说明无环存在
p1 = p2 = head; if(p2==NULL || p2->next==NULL)
return NULL;
p1 = p1->next;
p2 = p2->next->next; while(p1!=p2)
{
if(p2==NULL || p2->next==NULL)
return NULL;
p1 = p1->next;
p2 = p2->next->next; }
//此时p1==p2,说明有环存在。利用定理:p1和p2相遇处距离链表的第一个节点的长度是环长度的整数倍,来求环起始点。让一个指针从head处开始出发,另一个指针从相遇处出发,这两个指针必然在环起始节点处相遇
ListNode * CircleBegin_Node;
p1 = head;
while(p1!=p2)
{
p1 = p1->next;
p2 = p2->next;
}
CircleBegin_Node = p1;
return CircleBegin_Node;
}
};

[LeetCode OJ] Linked List Cycle II—Given a linked list, return the node where the cycle begins. If there is no cycle, return null.的更多相关文章

  1. 【LeetCode OJ】Palindrome Partitioning II

    Problem Link: http://oj.leetcode.com/problems/palindrome-partitioning-ii/ We solve this problem by u ...

  2. 【LeetCode OJ】Path Sum II

    Problem Link: http://oj.leetcode.com/problems/path-sum-ii/ The basic idea here is same to that of Pa ...

  3. 【LeetCode OJ】Word Ladder II

    Problem Link: http://oj.leetcode.com/problems/word-ladder-ii/ Basically, this problem is same to Wor ...

  4. 【LEETCODE OJ】Single Number II

    Problem link: http://oj.leetcode.com/problems/single-number-ii/ The problem seems like the Single Nu ...

  5. 【LeetCode OJ】Word Break II

    Problem link: http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the w ...

  6. LeetCode OJ——Pascal's Triangle II

    http://oj.leetcode.com/problems/pascals-triangle-ii/ 杨辉三角2,比杨辉三角要求的空间上限制Could you optimize your algo ...

  7. LeetCode OJ 45. Jump Game II

    Given an array of non-negative integers, you are initially positioned at the first index of the arra ...

  8. LeetCode OJ 229. Majority Element II

    Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorit ...

  9. LeetCode OJ 63. Unique Paths II

    Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...

随机推荐

  1. 【HDOJ】2757 Ocean Currents

    简单BFS. /* 2757 */ #include <iostream> #include <queue> #include <cstdio> #include ...

  2. 【算法Everyday】第三日 KMP算法

    题目 你知道的. 分析 分析不来. 代码 void OutputArray(int* pArr, int iLen) { ; i < iLen; i++) { printf("%d\t ...

  3. 【转】深入了解android平台的jni---注册native函数

    原文网址:http://my.oschina.net/u/157503/blog/169041 注册native函数有两种方法:静态注册和动态注册. 1.静态注册方法 根据函数名找到对应的JNI函数: ...

  4. kafka安装与使用

    一.下载 下载地址: http://kafka.apache.org/downloads.html kafka目录结构 目录 说明 bin 操作kafka的可执行脚本,还包含windows下脚本 co ...

  5. Android中ListView分页加载数据

    public class MainActivity extends Activity { private ListView listView=null; //listview的数据填充器 privat ...

  6. [Locked] Count Univalue Subtrees

    Count Univalue Subtrees Given a binary tree, count the number of uni-value subtrees. A Uni-value sub ...

  7. [Hibernate] 基本增删查改

    本文记录,Java 应用通过 Hibernate 对数据库 MySQL 进行基本的增删改查操作,即CRUD. 本例子的目录结构如下 hibernate.cfg.xml 存储数据库信息,如数据库类型,账 ...

  8. lightoj 1198 最大权重匹配

    题目链接:http://lightoj.com/volume_showproblem.php?problem=1198 #include <cstdio> #include <cst ...

  9. 356. Line Reflection

    首先找到X方向的中点,如果中点是一个点,那么分别从这个点开始往左右找就行:如果是一个区间,比如1 2之间,那么首先总点数得是偶数,然后以1和2往左右两边找就行.. 找的时候,有3种情况: 同时没找到, ...

  10. 利用spring AOP 实现统一校验

    开发环境 JDK: 1.7 spring: 4.0.6 aspect: 1.7.4 应用背景   在APP与后台通讯的过程中,我们一般都会有个authToken的字符串校验,判断那些请求是需要校验用户 ...