225. Find Node in Linked List Find a node with given value in a linked list. Return null if not exists. Example Example 1: Input: 1->2->3 and value = 3 Output: The last node. Example 2: Input: 1->2->3 and value = 4 Output: null Notice If there…

1.Remove Linked List Elements package linkedlist; /* * Question: Remove all elements from a linked list of integers that have value val. */ public class RemoveLinkedListElements { /** * @param head a ListNode * @param val an integer * @return a ListN…

Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note:Given n…

We are given a linked list with head as the first node.  Let's number the nodes in the list: node_1, node_2, node_3, ... etc. Each node may have a next larger value: for node_i, next_larger(node_i) is the node_j.val such that j > i, node_j.val > nod…

题目如下: We are given a linked list with head as the first node.  Let's number the nodes in the list: node_1, node_2, node_3, ... etc. Each node may have a next larger value: for node_i, next_larger(node_i) is the node_j.val such that j > i, node_j.val…

题目标签:Linked List, Stack 题目给了我们一个 Linked List,让我们找出对于每一个数字,它的下一个更大的数字. 首先把 Linked List 里的数字 存入 ArrayList, 方便后面的操作. 然后遍历 ArrayList,首先每一个数字,都会存入stack:所以就可以利用stack回到之前的数字,存入它的 next Greater Node. Java Solution: Runtime:  39 ms, faster than 65 % Memory Usa…

""" We are given a linked list with head as the first node. Let's number the nodes in the list: node_1, node_2, node_3, ... etc. Each node may have a next larger value: for node_i, next_larger(node_i) is the node_j.val such that j > i, n…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 单调递减栈 日期 题目地址:https://leetcode.com/problems/next-greater-node-in-linked-list/ 题目描述 We are given a linked list with head as the first node. Let's number the nodes in the list: no…

单调栈的应用. class Solution: def nextLargerNodes(self, head: ListNode) -> List[int]: stack = [] ret = [] while head: while stack and stack[-1][1] < head.val: ret[stack.pop()[0]] = head.val stack.append((len(ret), head.val)) ret.append(0) head = head.next…

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Given linked list -- head = [4,5,1,9], which looks like following: Example 1: Input: head = [4,5,1,9], node = 5 Output: [4,1,9] Explanation:…