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Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are non-negative.思路：通过一位乘法和字符串加法实现，注意进位问题，注意字符串方向问题。

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.For example, Given [0,1,0,2,1,0,1,3,2,1,2,1]

Given an unsorted integer array, find the first missing positive integer.For example,Given [1,2,0] return 3,and [3,4,-1,1] return 2.Your algorithm should run in O(n) time and uses constant

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.Each number in C may only be used once in the combina

1. 使用BlockingQueue 实现 生产者 消费者2. 使用 Semaphore 实现 多线程 依次执行3. 不是同一个对象，synchronized（）不能互斥4. "1"+""    编译器优化为  "1"  5.  不定期更新，记录

ConcurrentHashMapCopyOnWriteArrayListCopyOnWriteArraySet用法跟普通Map 、 List 、Set 一样，但是这些支持边迭代边修改等并发操作。

BlockingQueue最终会有四种状况，抛出异常、返回特殊值、阻塞、超时，下表总结了这些方法： 抛出异常特殊值阻塞超时插入add(e)offer(e)put(e)offer(e, time, unit)移除remove()poll()take()poll(time, unit)检查

Exchanger可以在两个线程之间交换数据，只能是2个线程，他不支持更多的线程之间互换数据。当线程A调用Exchange对象的exchange()方法后，他会陷入阻塞状态，直到线程B也调用了exchange()方法，然后以线程安全的方式交换数据，之后线程A和B继续运行package thread;import java.util.ArrayList;import java.

package thread;import java.text.SimpleDateFormat;import java.util.Date;import java.util.concurrent.CountDownLatch;public class CountDownLatchDemo { final static SimpleDateFormat sdf=new Simple

package thread;import java.io.IOException;import java.util.Random;import java.util.concurrent.BrokenBarrierException;import java.util.concurrent.CyclicBarrier;import java.util.concurrent.Executo

package thread;import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors;import java.util.concurrent.Semaphore;public class TestSemaphore { public static void main(St

package thread;import java.util.concurrent.locks.Condition;import java.util.concurrent.locks.Lock;import java.util.concurrent.locks.ReentrantLock;public class LockConditionTest { private Lock

# add-apt-repository ppa:webupd8team/java按下Enter ,导入密钥# apt-get update# apt-get install oracle-java8-installer# apt-get install oracle-java8-set-default默认安装路径：/usr/lib/jvm ，可根据需要配置环境变量。

HashMap 类中包含3个和扩容相关的常量：static final int DEFAULT_INITIAL_CAPACITY = 1 static final int MAXIMUM_CAPACITY = 1 static final float DEFAULT_LOAD_FACTOR = 0.75f;其中，DEFAULT_INITIAL_CAPACITY 是初始容量，默认

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.The same repeated number may be chosen from C unlimited numb

import java.util.HashMap;import java.util.Map;import java.util.Random;import java.util.concurrent.locks.ReadWriteLock;import java.util.concurrent.locks.ReentrantReadWriteLock;public class CacheS

代码前加锁，代码后解锁，为了保证能够解锁，通常用try-finally语句包起来。import java.util.concurrent.locks.Lock;import java.util.concurrent.locks.ReentrantLock;public class LockTest { private Lock lock = new ReentrantLock

import java.util.Random;import java.util.concurrent.Callable;import java.util.concurrent.CompletionService;import java.util.concurrent.ExecutionException;import java.util.concurrent.ExecutorComple

import java.util.Random;import java.util.concurrent.Callable;import java.util.concurrent.ExecutionException;import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors;impor

import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors;public class ThreadPool { public static void main(String[] args) { // 创建一个包含固定数量线程的线程池 ExecutorService th

1. 操作相同时，写一个 Runnable 实现类，内部设置成员变量，run 方法修改该变量，将该Runnable传给不同Thread使用；2. 操作不同时，在Thread调用类实例化一个数据实例，传递给不同Runnable处理，再把不同的Runnable传给不同Thread使用；3. 在Thread调用类实例化一个数据实例，使用内部Runnable类直接访问该数据实例进行处理，再把Run

package thread;import java.util.Random;public class ThreadLocalShareDataDemo { /** * ThreadLocal类及应用技巧 将线程范围内共享数据进行封装，封装到一个单独的数据类中，提供设置获取方法 * 将该类单例化，提供获取实例对象的方法，获取到的实例对象是已经封装好的当前线程范围内的对象 *

public class S03 { private boolean isSub = true; public synchronized void sub() throws InterruptedException { if (!isSub) { this.wait(); } // do something isSub = !isSub; this.notify(

静态方法 -->   锁是 类一般方法 -->  锁是  对象代码段  -->  锁是 自定义任何内容当锁相同时，方法间就能达到互斥效果。典型的3种加锁方式：public class S02 { public synchronized void out1(String str) { // something to do } public synch

new Timer().schedule(new TimerTask() { @Override public void run() { // TODO Auto-generated method stub } }, 5000);

例子代码： new Thread(new Runnable() { @Override public void run() { // TODO Auto-generated method stub } }) { @Override public void run() { // TODO Auto-generated method stub

The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 111221, ...1 is read off as "one 1" or 11.11 is read off as "two 1s" or 21.21 is read off as 

Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells are indicated by the character '.'.You may assume that there will be only one unique solution.A sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.The Sudoku board could be partially filled, where empty cells are filled with the character '.'.A partially fille

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You may assume no duplicates in the array.

Given a sorted array of integers, find the starting and ending position of a given target value.Your algorithm's runtime complexity must be in the order of O(log n).If the target is not found

Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).You are given a target value to search. If found in the array retur

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.For "(()", the longest valid parentheses substring is "()",

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.If such arrangement is not possible, it must rearrange it as the lowest possible

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in wordsexactly once and w

Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.思路：注意使用long进行计算，防止int溢出，然后再判断溢出时重置。public class Solution { public int divide(

Implement strStr().Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.思路：简单遍历查找即可。public class Solution { public int strStr(String ha

Given an array and a value, remove all instances of that value in place and return the new length.The order of elements can be changed. It doesn't matter what you leave beyond the new length.思

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.Do not allocate extra space for another array, you must do this in place with

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is