141. Linked List Cycle【easy】】的更多相关文章

141. Linked List Cycle[easy] Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x…
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 pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…
203. Remove Linked List Elements[easy] Remove all elements from a linked list of integers that have value val. ExampleGiven: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6Return: 1 --> 2 --> 3 --> 4 --> 5 Credits:Special than…
Remove all elements from a linked list of integers that have value val. Example: Input: 1->2->6->3->4->5->6, val = 6 Output: 1->2->3->4->5 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNo…
Given a linked list, determine if it has a cycle in it.   Example Given -21->10->4->5, tail connects to node index 1, return true Challenge Follow up:Can you solve it without using extra space? 解法一: class Solution { public: /** * @param head: The…
141. Linked List Cycle(Easy)2019.7.10 题目地址https://leetcode.com/problems/linked-list-cycle/ 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 pos which represents the position (0-i…
142. Linked List Cycle II[easy] 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? 解法一: /** * Definition for singl…
引入 快慢指针经常用于链表(linked list)中环(Cycle)相关的问题.LeetCode中对应题目分别是: 141. Linked List Cycle 判断linked list中是否有环 142. Linked List Cycle II 找到环的起始节点(entry node)位置. 简介 快指针(fast pointer)和慢指针(slow pointer)都从链表的head出发. slow pointer每次移动一格,而快指针每次移动两格. 如果快慢指针能相遇,则证明链表中有…
160. Intersection of Two Linked Lists[easy] Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 begin to intersect at n…
206. Reverse Linked List[easy] Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 解法一: class Solution { public: ListNode* reverseList(ListNode* head) { ListNode * pre = NULL;…
237. Delete Node in a Linked List[easy] Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the l…
234. Palindrome Linked List[easy] Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * Lis…
141. Sqrt(x) [easy] Implement int sqrt(int x). Compute and return the square root of x. Example sqrt(3) = 1 sqrt(4) = 2 sqrt(5) = 2 sqrt(10) = 3 Challenge  O(log(x)) 解法一: class Solution { public: /* * @param x: An integer * @return: The sqrt of x */…
83. Remove Duplicates from Sorted List[easy] Given a sorted linked list, delete all duplicates such that each element appear only once. For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3. 解法一: /** * Defin…
21. Merge Two Sorted Lists[easy] 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. 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val…
题目: 141.Given a linked list, determine if it has a cycle in it. 142.Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 思路: 带环链表如图所示.设置一个快指针和一个慢指针,快指针一次走两步,慢指针一次走一步.快指针先进入环,慢指针后进入环.在进入环后,可以理解为快指针追赶慢指针,由于两个指…
判断链表有环,环的入口结点,环的长度 1.判断有环: 快慢指针,一个移动一次,一个移动两次 2.环的入口结点: 相遇的结点不一定是入口节点,所以y表示入口节点到相遇节点的距离 n是环的个数 w + n + y = 2 (w + y) 经过化简,我们可以得到:w  = n - y; https://www.cnblogs.com/zhuzhenwei918/p/7491892.html 3.环的长度: 从入口结点或者相遇的结点移动到下一次再碰到这个结点计数 https://blog.csdn.ne…
141. 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? 利用快慢指针,如果相遇则证明有环 注意边界条件: 如果只有一个node. public class Solution { public boolean hasCycle(ListNode head) { if(head==null |…
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