【LeetCode刷题】最长同值路径：妙解

这题主要学习巧用C++语言,内联函数.结构体排序.C++书写方式 static int fast_streams = []() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); std::cout.tie(nullptr); return 0; }(); struct Int2 { int x, y; }; inline bool operator==(Int2 a, Int2 b) { return a.x == b.x &&a

题目描述 编写一个函数来查找字符串数组中的最长公共前缀. 如果不存在公共前缀,返回空字符串 "". 示例 1: 输入: ["flower","flow","flight"]输出: "fl" 示例 2: 输入: ["dog","racecar","car"]输出: ""解释: 输入不存在公共前缀. 说明: 所有输入只包含小写字母 

最长同值路径 给定一个二叉树,找到最长的路径,这个路径中的每个节点具有相同值. 这条路径可以经过也可以不经过根节点. 注意:两个节点之间的路径长度由它们之间的边数表示. 示例 1: 输入: 输出: 2 示例 2: 输入: 输出: 2 注意: 给定的二叉树不超过10000个结点. 树的高度不超过1000. 思路 Intuition We can think of any path (of nodes with the same values) as up to two arrows extendi

Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example,Given [100, 4, 200, 1, 3, 2],The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. Your algorithm should run in

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / 4 8 / / 11 13 4 / \ / 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 给定一个二叉树和数字sum,输出二叉树中从根节点到

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum of all root-to-leaf numbers. For example, 1 / 2 3 T

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 给出二叉树的中序遍历和后序遍历结果,恢复出二叉树. 后序遍历序列的最后一个元素值是二叉树的根节点的值,查找该元素在中序遍历序列中的位置mid,根据中序遍历和后序遍历性质,有: 位置mid以前的序列部分为二叉树根节点左子树中

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / 4 8 / / 11 13 4 / \ 7 2 1 return true, as t

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 题解: 思路比较简单,每条直线都可以表示为y=kx+b,所以对于任意三点,如果它们共线,那么它们中任意两点的斜率都相等. 所以就遍历points数组,对其中的每一个元素计算它和位于它后面的数组元素的斜率并保存在一个hashmap中. 这个hashmap的键就是两点构成直线的斜率,值就是和当前元素po

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys