LeetCode & linked list bug add-two-numbers shit test /** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */ var addTwo…

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? 和问题一Linked List Cycle几乎一样.如果用我的之前的解法的话,可以很小修改就可以实现这道算法了.但是如果问题一用优化了的解法的话,那么就不适…

1.Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. 使用hash public int[] twoSum…

Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen. Follow up: What if the linked list is extremely large and its length is unknown to you? Could you solve this effi…

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? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 这道题是快慢指针的经典应用.只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇.实在是太巧妙了,要是我肯定想不出来.代码如下: C++ 解法: class Solution { public: bool hasCycle…

问题描述如下: 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? 从问题来看,如果可以充分利用额外空间的话,这个题目是不难的,然而题目提出了一个要求,能否在不使用任何额外空间的情况下解决这个问题. 通过反复思考,我觉得这题类似于追击问题,可以用一个…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 原题链接:https://oj.leetcode.com/problems/linked-list-cycle/ 题目:给定一个链表.推断它是否有环. 继续: 你能不用额外的空间解决吗? 思路:使用两个指针fast,slow,fast每次向前走两步,slow每次向前走一步.假设…

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? 原题链接:https://oj.leetcode.com/problems/linked-list-cycle-ii/ 题目:给定一个链表.返回环開始的节点.如无环.返回null. public L…

We are given head, the head node of a linked list containing unique integer values. We are also given the list G, a subset of the values in the linked list. Return the number of connected components in G, where two values are connected if they appear…

题目要求 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 如何判断一个单链表中有环? Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle…

这题参照http://blog.jobbole.com/42550/ 用的蓄水池算法,即更改ans的概率为1/(当前length) /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { private ListNode head; /**…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? Linked List Two Pointers     ''' Created on Nov 13, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com> ''' # De…

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? 借用博客http://www.cnblogs.com/hiddenfox/p/3408931.html的图 设环的距离为L = (b+c),无环的距离为a, 假设在时间t相遇,慢指针行驶距离为x,则…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 求链表是否有环的问题,要考虑链表为空的情况,定义一个快指针和一个慢指针,如果快指针和慢指针重合就表明有环 bool hasCycle(ListNode *head){ if(head == NULL || head->next == NULL) return false; Lis…

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. For example, Given 1->2->3->3->4->4->5, return 1->2->5. Given 1->1->1->2->3, return 2->3…

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? SOLUTION 1: 1. 先用快慢指针判断是不是存在环. 2. 再把slow放回Start处,一起移动,直到二个节点相遇,就是交点.…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? SOLUTION 1: 经典快慢指针问题.如果存在环,fast, slow必然会相遇.就像2个速度不一样的人在环形跑道赛跑,总有一个时间他们会相遇. 如果fast到达了null,就是没有环,可以返回false. package Algorith…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 错误解法.因为Cycle可能出现在Link的中间的,所以需要检查其中间Nodes. bool hasCycle(ListNode *head) { // IMPORTANT: Please reset any member data you…

Problem Description Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? Problem Solution /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(in…

[题目] Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? [题意] 推断一个单向链表是否有环 [思路] 维护两个指针p1和p2,p1每次向前移动一步,p2每次向前移动两步     假设p2可以追上p1,则说明链表中存在环 [代码] /** * Definition for singly-linked list. * stru…

Question: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Analysis: 只想到了,首先判断是否有环,若有环,则总链首开始,一次判断是否是环的开始,这样T(O) = O(n^2). 其实是一个数学问题,详细思路参照链接http://blog.csdn.net/sbitswc/article/details/27584037 . Answer…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断链表中是否有环,不能用额外的空间,可以使用快慢指针,慢指针一次走一步,快指针一次走两步,若是有环则快慢指针会相遇,若是fast->next==NULL则没有环. 值得注意的是:在链表的题中,快慢指针的使用频率还是很高,值得注意. /** * Definition for si…

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up:Can you solve it without using extra space? 题意:给定链表,若是有环,则返回环开始的节点,没有则返回NULL 思路:题目分两步走,第一.判断是否有环,第二若是有,找到环的起始点.关于第一点,可以参考之前的博客 Linked list cycle.…

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? Hide Tags Linked List Two Pointers        开始犯二了一点,使用了node 的val 作为判断,其实是根据内存判断.找出链表环的起始位置,这个画一下慢慢找下规律…

问题描述: Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycl…

Question : 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? Anaylsis : 首先,比较直观的是,先使用Linked List Cycle I的办法,判断是否有cycle.如果有,则从头遍历节点,对于每一个节点,查询是否在环里面,是…

Description Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1: Input: ->-> Output: -> Example 2: Input: ->->->-> Output: ->-> 问题描述:给定一个已排序链表,移除重复元素 代码: public ListNode DeleteDup…

Description Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: ->->, ->-> Output: ->->->->-> 问题描述,将两个排序的链表归并 代码: public L…

Description You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the…