mengq303的专栏

leetcode50题String to Integer的实现

给定一个数组，左移k位。这一题，我开始是这样考虑的，每一位都应该直接移动到 (i-k+n)%n位。所以代码如下：import java.util.Scanner;public class Reverse { public String solve(char[] c, int k) { int i = 0; char temp = c[i]; int n = c.

Given an array of integers, every element appears three times except for one. Find that single one.这一题好难啊。一点头绪都没有，要求是线性时间复杂度，同时要求常数项空间。答案上，是这样来的。使用了位运算。ps：感觉用到位运算的题目都好难。如果对所有数字的第i位bit进行统计的话，会发

Given a binary tree, find the maximum path sum.For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The

Given an array of integers, every element appears twice except for one. Find that single one.这一题第一反应是 直接存个map，作为键值对。但是这一题要求 不要有extra memory，同时时间复杂度为O(n)；直接想到异或^，两个数相异为1，相同为0。所以把数组整个异或一遍，出来的值就是

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the origin

Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.一开始的考虑很有问题，忘了考虑是到结点（左右子树都为空）的路径，所以

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.和上一题不同的地方就在于这一题是sorted list 上一题是 sorted array比较无耻点，其实可以直接仿照之前的sorted array的方法：代码如下（有问

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

Given a binary tree, determine if it is height-balanced.For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

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.Th

Given a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.这一题很简单，就什么都不多说了。数的高度等于子树的高度（

题目：A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.Return a deep copy of the list.不要想的太复杂了。直接用个hash

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.这题一开始一直TLE。后来查了下，发现是因为/** * Definition for singly-linked list. * public class ListNode { *