10 2022 档案
摘要：Sort Integers by The Number Of 1 Bits You are given an integer array arr. Sort the integers in the array in ascending order by the number of 1's in th
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摘要：Keyboard Row Given an array of strings words, return the words that can be typed using letters of the alphabet on only one row of American keyboard li
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摘要：First Bad Version You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product f
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摘要：Missing Number Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the arr
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摘要：Number Of 1 Bits Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight). Note: N
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摘要：Two Sum Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that
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摘要：Calculate Money in Leetcode Bank Hercy wants to save money for his first car. He puts money in the Leetcode bank every day. He starts by putting in $1
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摘要：Add to Array-Form of Integer The array-form of an integer num is an array representing its digits in left to right order. For example, for num = 1321,
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摘要：Check if One String Swap Can Make Strings Equal You are given two strings s1 and s2 of equal length. A string swap is an operation where you choose tw
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摘要：Minimum Sum of Four Digit Number After Splitting Digits You are given a positive integer num consisting of exactly four digits. Split num into two new
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摘要：Sum of Unique Elements You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array. Re
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摘要：in ing y 因姻殷音阴 银龈吟寅淫 引蚓隐瘾饮尹 印荫 英应鹰婴樱缨鹦 营莹萤盈迎赢 影 映硬应 b 宾滨缤彬 殡鬓 兵冰丙柄秉饼禀 病并 p 拼 贫频 品 聘 乒平苹萍屏瓶凭 m 民 敏皿闽悯泯 名茗铭明鸣冥命 d 丁叮钉仃叮 顶鼎定锭订 t 听厅汀 亭停廷庭蜒 挺艇 n 您 宁狞拧凝 拧
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摘要：Largest Perimeter Triangle Given an integer array nums, return the largest perimeter of a triangle with a non-zero area, formed from three of these le
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摘要：Consecutive Characters The power of the string is the maximum length of a non-empty substring that contains only one unique character. Given a string
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摘要：Valid Palindrome II Given a string s, return true if the s can be palindrome after deleting at most one character from it. Example 1: Input: s = "aba"
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摘要：sz, rz 命令能在 ssh 连接的时候快速传输文件 sz (send 发送) rz (receive 接收) 使用这两个命令的前提是服务端安装了命令，没有安装的可以用下面的命令安装 yum install lrzsz 并且 ssh 连接工具也要支持，putty 本身是不支持的，但是我们可以下载
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摘要：Occurrences After Bigram second third", where second comes immediately after first, and third comes immediately after second. Return an array of all t
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摘要：Remove Palindromic Subsequences You are given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subse
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摘要：Count Good Triplets Given an array of integers arr, and three integers a, b and c. You need to find the number of good triplets. A triplet (arr[i], ar
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摘要：Binary Search Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If
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摘要：Minimum Changes To Make Alternatign Binary String You are given a string s consisting only of the characters '0' and '1'. In one operation, you can ch
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摘要：一个 Demo，记录一下发布公众号图文时需要用到的接口 公众号开发时需用到的一些网站 微信官方文档平台 ,开发公众号只用查看公众号那一块 微信公众平台接口测试帐号申请 ,申请公众号测试账号 微信公众平台接口调试工具 ,官方提供的接口测试工具 WxJava 仓库 , WxJava仓库地址 natapp
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摘要：Minimum Hours of Training to Win a Competition You are entering a competition, and are given two positive integers initialEnergy and initialExperience
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摘要：Minimum Index Sum of Two Lists Given two arrays of strings list1 and list2, find the common strings with the least index sum. A common string is a str
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摘要：Number Complement The complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representa
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摘要：Decompress Run-length Encoded List We are given a list nums of integers representing a list compressed with run-length encoding. Consider each adjacen
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摘要：NimGame You are playing the following Nim Game with your friend: Initially, there is a heap of stones on the table. You and your friend will alternate
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摘要：Self Dividing Numbers A self-dividing number is a number that is divisible by every digit it contains. For example, 128 is a self-dividing number beca
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摘要：Is Subsequence Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence of a string is a new string that i
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摘要：Make Array Zero by Subtracting Equal Amounts You are given a non-negative integer array nums. In one operation, you must: Choose a positive integer x
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摘要：Find Closest Number To Zero Given an integer array nums of size n, return the number with the value closest to 0 in nums. If there are multiple answer
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摘要：Summary Ranges You are given a sorted unique integer array nums. A range [a,b] is the set of all integers from a to b (inclusive). Return the smallest
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摘要：Counting Bits Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary re
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摘要：Word Pattern Given a pattern and a string s, find if s follows the same pattern. Here follow means a full match, such that there is a bijection betwee
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摘要：Add Digits Given an integer num, repeatedly add all its digits until the result has only one digit, and return it. Example 1: Input: num = 38 Output:
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摘要：somorphic Strings Given two strings s and t, determine if they are isomorphic. Two strings s and t are isomorphic if the characters in s can be replac
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摘要：Reverse String II Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string. If
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摘要：Base 7 Given an integer num, return a string of its base 7 representation. Example 1: Input: num = 100 Output: "202" Example 2: Input: num = -7 Output
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摘要：Fibonacci Number The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the
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摘要：Ransom Note Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false other
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摘要：Sum of Left Leaves Given the root of a binary tree, return the sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that i
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摘要：Assign Cookies Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each
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摘要：Fizz Buzz Given an integer n, return a string array answer (1-indexed) where: answer[i] == "FizzBuzz" if i is divisible by 3 and 5. answer[i] == "Fizz
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摘要：Valid Anagram Given two strings s and t, return true if t is an anagram of s, and false otherwise. An Anagram is a word or phrase formed by rearrangin
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摘要：Remove Linked List Elements Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and
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摘要：Move Zeroes 思路一: 用 left 指针标记求解数组最大下标 + 1，初始化的时候是 0，随着往右遍历，left 会一步步扩大。最后把 left 右边的数都置为 0。 这题的关键在于 left 永远指向求解数组最大下标 + 1 public void moveZeroes(int[] n
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摘要：Power Of Two 思路一: 观察 2 的 n 次方的二进制，都只有一位 1 bit，遍历即可 public boolean isPowerOfTwo(int n) { if (n <= 0) return false; int count = 0; for (int i = 0; i < 3
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摘要：Majority Element 思路一: map public int majorityElement(int[] nums) { if (nums.length == 1) return nums[0]; Map<Integer, Integer> map = new HashMap<>();
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摘要：Reverse Bits 思路一: 遍历 32 位 bit，记录 bit 结果 public int reverseBits(int n) { int result = 0; int x = 32; while (x-- > 0) { int bit = n & 1; result <<= 1; i
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摘要：下面操作是在虚拟机环境下，根目录最好不要扩容或者缩小，一旦出问题很容易导致系统启动分区故障 查看磁盘信息fdisk -l Disk /dev/sda: 48 GiB, 51539607552 bytes, 100663296 sectors ... Device Boot Start End Sec
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摘要：Long Pressed Name 思路一: 暴力，遍历两个字符串，对比。边界情况不太好处理 public boolean isLongPressedName(String name, String typed) { if (typed.length() < name.length()) retur
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摘要：X of a Kind in a Deck of Cards 思路一: 统计数字个数，然后用 [2, length] 的约数尝试所有数字个数，判断能否整除 public boolean hasGroupsSizeX(int[] deck) { Map<Integer, Integer> map =
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摘要：Monotonic Array 思路一: 遍历 public boolean isMonotonic(int[] nums) { boolean increase = true; for (int i = 0; i < nums.length - 1; i++) { if (nums[i] > nu
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摘要：Binary Gap 思路一: 记录 bit 为 1 的位置，保留区间最大的值 public int binaryGap(int n) { int idx = -1; int result = 0; int i = 0; while (n > 0) { if ((n & 1) == 1) { if
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摘要：Buddy Strings 思路一: 两个字符是 Buddy Strings 的完全分类 字符不一样 && 只存在两个位置交换的字符 字符一样 && 存在重复的字符 按照上面的分类，记录两个不同位置的下标和统计出现字符的情况 public boolean buddyStrings(String s,
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摘要：Flipping an Image 思路一: 遍历数组，先对数组进行 flip 然后再对数组进行 invert public int[][] flipAndInvertImage(int[][] image) { int N = image.length; for (int i = 0; i < N
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摘要：Teemo Attacking 思路一: 对两个前后攻击序列，完全分类只可能出现两种情况，只要把重叠的时间都减去，就是所求时间。此分类对三个以上的重叠时间依旧成立 没有重叠，攻击时间累加 有重叠，攻击时间需要去除重叠部分 public int findPoisonedDuration(int[] t
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摘要：Hamming Distance 思路一: 求 bit 不同的位数，XOR 可以把 bit 不同的位置变成 1，再用 bitCount() 统计 public int hammingDistance(int x, int y) { return Integer.bitCount(x ^ y); }
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摘要：思路一: 用数组记录 public List<Integer> findDisappearedNumbers(int[] nums) { int[] m = new int[nums.length]; for (int num : nums) { if (num-1 <= m.length) { m
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摘要：问题: 想看看容器内 java 线程锁的状态，运行 jstack 1 提示 “Unable to get pid of LinuxThreads manager thread”，使用其他 jdk 命令也是如此 排查：原因是因为容器内的 java pid 为 1，导致内置的 jdk 命令无法使用，有个
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摘要：Find the Difference 思路一: xor 两个字符串 public char findTheDifference(String s, String t) { char result = 0; for (int i = 0; i < s.length(); i++) { result
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摘要：Invert Binary Tree 思路一：递归，交换左右。这题比较出名，个人感觉面试的题目和实际工作中遇到的问题还是不太一样的，所以一点准备都不做就跑去面试，答不上来很正常。 一般能力强的人过一遍资料就有大致的知识结构了 public TreeNode invertTree(TreeNode r
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摘要：Contains Duplicate II 思路一: for 循环遍历，结果超时 public boolean containsNearbyDuplicate(int[] nums, int k) { int left = -1; for (int i = 0; i < nums.length; i
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摘要：Contains Duplicate 思路一: Set 检测 public static boolean containsDuplicate(int[] nums) { Set<Integer> set = new HashSet<>(); for (int num : nums) { if (se
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摘要：To Lower Case 思路一: 遍历，遇到 A-Z 范围内的 char 转换到对应 a-z 范围 public String toLowerCase(String s) { if (s == null || s.isEmpty()) return s; int gap = 'a' - 'A';
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摘要：Add String 思路一: 模拟加法运算，字符串前面填零 public String addStrings(String num1, String num2) { int max = Math.max(num1.length(), num2.length()); num1 = pad(num1,
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摘要：Reverse String 思路一: 首尾互换 public void reverseString(char[] s) { int mid = s.length / 2; int begin = 0; int end = s.length - 1; while (begin < mid) { ch
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摘要：Ugly Number 思路一: 从数字中依次去除 2，3，5，查看剩余的值是否为 1 public boolean isUgly(int n) { if (n == 0) return false; while (n % 2 == 0) { n /= 2; } while (n % 3 == 0)
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摘要：Happy Number 思路一: happy number 的结果完全分类，就两种情况 最后的值为 1 进入循环(用 map 记录) public boolean isHappy(int n) { Set<Integer> set = new HashSet<>(); while (true) {
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摘要：Linked List Cycle 思路一: 用 set 记录每个节点的 hashCode，如果遇到重复，说明是循环 public boolean hasCycle(ListNode head) { Set<Integer> set = new HashSet<>(); while (head !=
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摘要：Single Number 思路一: 用 set 过滤，剩下唯一一个就是目标数字 public int singleNumber(int[] nums) { Set<Integer> set = new HashSet<>(); for (int num : nums) { if (set.cont
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摘要：Maximum Depth of Binary Tree 思路一: 递归 public int maxDepth(TreeNode root) { if (root == null) return 0; return 1 + Math.max(maxDepth(root.left), maxDept
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摘要：Symmetric Tree 思路一: 递归 public boolean isSymmetric(TreeNode left, TreeNode right) { if (left == null && right == null) return true; if (left == null ||
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摘要：Same Tree public boolean isSameTree(TreeNode p, TreeNode q) { if (p == null && q == null) return true; if (p != null && q == null) return false; if (p
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摘要：Binary Tree Inorder Traversal public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<>(); t(root, list); return lis
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摘要：Merge Sorted Array 思路一: 比较两个数组前面最小值，依次插入到新数组中，最后复制新数组到 num1 中 public void merge(int[] nums1, int m, int[] nums2, int n) { int[] result = new int[nums1
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摘要：Remove Duplicates from Sorted List 思路一: 双指针,左指针记录链表最后有效位置，右指针向前扫描，遇到不重复的值，加到左指针后面，双指针依次向前 public ListNode deleteDuplicates(ListNode head) { if (head =
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摘要：Climbing Stairs 思路一: 动态规划，假设爬上第 n 阶楼梯，完全分类只可能存在两种情况 在 n-1 楼梯处直接一步上来 在 n-2 楼梯处直接两步上来 所以 爬上第 n 阶楼梯的方法: f(n) = f(n-1) + f(n+1) public int climbStairs(int
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摘要：Sqrt(x) 思路一: 暴力 public int mySqrt(int x) { long begin = 1L; while ((begin * begin) <= x) { begin++; } return Long.valueOf(begin).intValue() - 1; } 思路二
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摘要：问题: docker-compose 启动 java 容器时报错 library initialization failed - unable to allocate file descriptor table - out of memoryPicked up JAVA_TOOL_OPTIONS:
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摘要：Add Binary 思路一: 先计算公共部分，最后补充未计算的位置，模拟二进制加法，写的太丑了 public String addBinary(String a, String b) { char ONE = '0' + '1'; char TWO = '1' + '1'; StringBuild
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摘要：Plus One 思路一: 暴力，方向想错了，不能把 digits 当做一个整数看 public int[] plusOne(int[] digits) { if (digits[digits.length - 1] != 9) { digits[digits.length - 1]++; retu
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摘要：Length of Last Word 思路一: 从后面非空格字符开始扫描，记录非空格字符个数。优化：不用 char[]，直接用 charAt() 判断 public int lengthOfLastWord(String s) { char[] chars = s.toCharArray(); i
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摘要：Remove Duplicates from Sorted Array 思路一: 双指针，左指针永远指向有效数组长度+1的位置，左指针只会在出现交换后向右移动。右指针一直向右扫描，遇到不重复的数字就和左指针交换。 public int removeDuplicates(int[] nums) { i
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摘要：一个简单的 java web 程序，运行在 docker 容器中，提供对外接口服务 现象：使用 docker run 启动容器后可以正常运行，访问一切正常,但是隔了几十分钟后容器自动退出，java 日志一切正常 排查： docker ps -a 查看挂掉的容器 id，docker logs 查看容器
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摘要：Search Insert Position 思路一: 二分查找，然后处理没有找到的情况 public int searchInsert(int[] nums, int target) { int i = Arrays.binarySearch(nums, target); if (i >= 0)
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摘要：Valid Parentheses 思路一: 用栈处理,括号匹配就出栈，如果最后栈不为空，说明输入的括号不匹配 public boolean isValid(String s) { char[] chars = s.toCharArray(); Deque<Character> stack = ne
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摘要：Longest Common Prefix 思路一: 取第一个字符串为基准，依次对比剩余的字符串，取公共串 private static int commonPrefix(char[] arr1, char[] arr2) { int n = Math.min(arr1.length, arr2.l
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摘要：remove element 思路一: 先排序，然后去除数字 public static int removeElement(int[] nums, int val) { Arrays.sort(nums); int INVALID = -1; int L = INVALID; for (int i
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摘要：merge two sorted lists 思路一: 暴力求解，依次把两链表元素插入到新链表中 public ListNode mergeTwoLists(ListNode list1, ListNode list2) { if (list1 == null && list2 == null) r
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摘要：Roman to Integer 思路一: 暴力求解，遍历字符，遇到 6 种特殊的组合字符单独计算 static Map<Character, Integer> map = new HashMap<>(); static Map<String, Integer> combine = new Hash
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摘要：回文数 思路一: 暴力求解，把数字一个一个拆分，放队列里面，最后取队列的首尾，对比是否相同 public static boolean isPalindrome2(int x) { if (x < 0) return false; if (x < 10) return true; Deque<Int
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摘要：思路一:暴力求解 public int[] twoSum(int[] nums, int target) { for(int i = 0; i < nums.length; i++) { for(int j = i+1; j < nums.length; j++) { if (nums[i] + n
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摘要：思路一：分别用两个 map 记录最大值和最小值，然后遍历这两个 map 求得最大距离 public int maxDistance(int[] colors) { Map<Integer, Integer> min = new HashMap<>(); Map<Integer, Integer> m
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