多线程的创建与常见使用方法
一、多线程创建方式
1. 继承Thread类
class MyThread extends Thread {
@Override
public void run() {
// 线程执行逻辑
System.out.println(Thread.currentThread().getName() + " is running");
}
}
// 使用
MyThread thread = new MyThread();
thread.start(); // 启动线程2. 实现Runnable接口
class MyRunnable implements Runnable {
@Override
public void run() {
System.out.println(Thread.currentThread().getName() + " is running");
}
}
// 使用
Thread thread = new Thread(new MyRunnable());
thread.start();3. Android特有方式
// HandlerThread（自带Looper）
HandlerThread handlerThread = new HandlerThread("WorkerThread");
handlerThread.start();
Handler handler = new Handler(handlerThread.getLooper()) {
@Override
public void handleMessage(Message msg) {
// 后台线程处理消息
}
};
// 发送消息
handler.sendEmptyMessage(0);二、线程池使用（推荐方式）
1. 创建线程池
// 固定大小线程池
ExecutorService fixedPool = Executors.newFixedThreadPool(4);
// 缓存线程池（自动扩容）
ExecutorService cachedPool = Executors.newCachedThreadPool();
// 单线程池（顺序执行）
ExecutorService singleThreadPool = Executors.newSingleThreadExecutor();2. 提交任务
// 执行Runnable任务
fixedPool.execute(() -> {
System.out.println("Task running in thread pool");
});
// 提交Callable任务并获取Future
Future<String> future = fixedPool.submit(() -> {
return "Task result";
});3. 自定义线程池
ThreadPoolExecutor customPool = new ThreadPoolExecutor(
4, // 核心线程数
10, // 最大线程数
60, // 空闲线程存活时间（秒）
TimeUnit.SECONDS,
new LinkedBlockingQueue<>(100) // 任务队列
);
// 拒绝策略（当队列满时）
customPool.setRejectedExecutionHandler((task, executor) -> {
System.err.println("Task rejected: " + task);
});
1. 交替打印数字（两个线程协作）
题目：两个线程交替打印1~100，一个线程打印奇数，另一个打印偶数
public class AlternatePrint {
private int num = 1;
private final Object lock = new Object();
public void printOdd() {
synchronized (lock) {
while (num <= 100) {
if (num % 2 == 1) {
System.out.println(Thread.currentThread().getName() + ": " + num++);
lock.notify(); // 唤醒偶数线程
} else {
try {
lock.wait(); // 等待偶数线程通知
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
}
public void printEven() {
synchronized (lock) {
while (num <= 100) {
if (num % 2 == 0) {
System.out.println(Thread.currentThread().getName() + ": " + num++);
lock.notify(); // 唤醒奇数线程
} else {
try {
lock.wait(); // 等待奇数线程通知
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
}
public static void main(String[] args) {
AlternatePrint printer = new AlternatePrint();
new Thread(printer::printOdd, "OddThread").start();
new Thread(printer::printEven, "EvenThread").start();
}
}2. 生产者-消费者模型（缓冲区管理）
题目：实现生产者线程生成数据，消费者线程消费数据，使用阻塞队列控制
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
public class ProducerConsumer {
private static final int CAPACITY = 5;
private final BlockingQueue<Integer> queue = new ArrayBlockingQueue<>(CAPACITY);
public void produce() throws InterruptedException {
int value = 0;
while (true) {
synchronized (this) {
while (queue.size() == CAPACITY) {
wait(); // 缓冲区满时等待
}
queue.put(value);
System.out.println("Produced: " + value);
value++;
notifyAll(); // 通知消费者
}
Thread.sleep(100); // 模拟生产耗时
}
}
public void consume() throws InterruptedException {
while (true) {
synchronized (this) {
while (queue.isEmpty()) {
wait(); // 缓冲区空时等待
}
int value = queue.take();
System.out.println("Consumed: " + value);
notifyAll(); // 通知生产者
}
Thread.sleep(150); // 模拟消费耗时
}
}
public static void main(String[] args) {
ProducerConsumer pc = new ProducerConsumer();
new Thread(() -> {
try { pc.produce(); }
catch (InterruptedException e) { Thread.currentThread().interrupt(); }
}, "Producer").start();
new Thread(() -> {
try { pc.consume(); }
catch (InterruptedException e) { Thread.currentThread().interrupt(); }
}, "Consumer").start();
}
}3. 多线程顺序打印（线程协同）
题目：三个线程按顺序循环打印ABC各10次
public class SequentialPrinter {
private final Object lock = new Object();
private String state = "A";
public void printA() {
synchronized (lock) {
for (int i = 0; i < 10; ) {
if (state.equals("A")) {
System.out.print("A");
state = "B";
i++;
lock.notifyAll(); // 唤醒所有等待线程
} else {
try {
lock.wait(); // 等待条件满足
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
}
// 类似实现printB()和printC()
public void printB() {
synchronized (lock) {
for (int i = 0; i < 10; ) {
if (state.equals("B")) {
System.out.print("B");
state = "C";
i++;
lock.notifyAll();
} else {
try {
lock.wait();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
}
public void printC() {
synchronized (lock) {
for (int i = 0; i < 10; ) {
if (state.equals("C")) {
System.out.print("C ");
state = "A";
i++;
lock.notifyAll();
} else {
try {
lock.wait();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
}
public static void main(String[] args) {
SequentialPrinter printer = new SequentialPrinter();
new Thread(printer::printA, "Thread-A").start();
new Thread(printer::printB, "Thread-B").start();
new Thread(printer::printC, "Thread-C").start();
}
}4. 多线程计算素数（任务分发）
题目：使用线程池计算1~N范围内素数的数量
import java.util.concurrent.*;
public class PrimeCounter {
public static void main(String[] args) throws Exception {
final int N = 1000000;
final int THREADS = Runtime.getRuntime().availableProcessors();
ExecutorService executor = Executors.newFixedThreadPool(THREADS);
int segment = N / THREADS;
Future<Integer>[] results = new Future[THREADS];
// 分发任务
for (int i = 0; i < THREADS; i++) {
final int start = i * segment + 1;
final int end = (i == THREADS - 1) ? N : (i + 1) * segment;
results[i] = executor.submit(() -> {
int count = 0;
for (int num = start; num <= end; num++) {
if (isPrime(num)) count++;
}
return count;
});
}
// 汇总结果
int totalPrimes = 0;
for (Future<Integer> future : results) {
totalPrimes += future.get();
}
System.out.println("Total primes: " + totalPrimes);
executor.shutdown();
}
private static boolean isPrime(int n) {
if (n <= 1) return false;
if (n == 2) return true;
if (n % 2 == 0) return false;
for (int i = 3; i <= Math.sqrt(n); i += 2) {
if (n % i == 0) return false;
}
return true;
}
}5. 自定义简单线程池
题目：实现一个固定大小的线程池，支持提交任务和执行任务
import java.util.concurrent.*;
import java.util.*;
public class SimpleThreadPool {
private final BlockingQueue<Runnable> taskQueue;
private final List<WorkerThread> workers;
private volatile boolean isShutdown = false;
public SimpleThreadPool(int poolSize) {
this.taskQueue = new LinkedBlockingQueue<>();
this.workers = new ArrayList<>();
for (int i = 0; i < poolSize; i++) {
WorkerThread worker = new WorkerThread("Worker-" + i);
worker.start();
workers.add(worker);
}
}
public void execute(Runnable task) {
if (isShutdown) {
throw new IllegalStateException("ThreadPool is shutdown");
}
taskQueue.offer(task);
}
public void shutdown() {
isShutdown = true;
workers.forEach(Thread::interrupt);
}
private class WorkerThread extends Thread {
public WorkerThread(String name) {
super(name);
}
@Override
public void run() {
while (!isShutdown || !taskQueue.isEmpty()) {
try {
Runnable task = taskQueue.take();
task.run();
} catch (InterruptedException e) {
// 处理中断，结束线程
}
}
}
}
public static void main(String[] args) {
SimpleThreadPool pool = new SimpleThreadPool(4);
// 提交任务
for (int i = 0; i < 10; i++) {
final int taskId = i;
pool.execute(() -> {
System.out.println(Thread.currentThread().getName()
+ " executing Task-" + taskId);
try {
Thread.sleep(500);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
});
}
// 等待任务完成
try {
Thread.sleep(3000);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
pool.shutdown();
}
}字符串算法总结
1. 字符串反转
题目：反转字符串或单词顺序
// 整个字符串反转
public String reverseString(String s) {
char[] chars = s.toCharArray();
int left = 0, right = chars.length - 1;
while (left < right) {
char temp = chars[left];
chars[left++] = chars[right];
chars[right--] = temp;
}
return new String(chars);
}
// 单词顺序反转（保留空格）
public String reverseWords(String s) {
String[] words = s.trim().split("\\s+");
StringBuilder sb = new StringBuilder();
for (int i = words.length - 1; i >= 0; i--) {
sb.append(words[i]);
if (i > 0) sb.append(" ");
}
return sb.toString();
}2. 字符串转换整数 (atoi)
题目：处理边界条件（空格/正负号/溢出）
public int myAtoi(String s) {
int index = 0, sign = 1, total = 0;
// 1. 跳过空格
while (index < s.length() && s.charAt(index) == ' ') index++;
if (index == s.length()) return 0;
// 2. 处理正负号
if (s.charAt(index) == '+' || s.charAt(index) == '-') {
sign = s.charAt(index) == '+' ? 1 : -1;
index++;
}
// 3. 转换数字并处理溢出
while (index < s.length()) {
int digit = s.charAt(index) - '0';
if (digit < 0 || digit > 9) break;
// 检查溢出
if (total > Integer.MAX_VALUE / 10 ||
(total == Integer.MAX_VALUE / 10 && digit > 7)) {
return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
}
total = total * 10 + digit;
index++;
}
return total * sign;
}3. 最长公共前缀
题目：查找字符串数组的最长公共前缀
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
for (int i = 0; i < strs[0].length(); i++) {
char c = strs[0].charAt(i);
for (int j = 1; j < strs.length; j++) {
if (i == strs[j].length() || strs[j].charAt(i) != c) {
return strs[0].substring(0, i);
}
}
}
return strs[0];
}4. 无重复字符的最长子串
题目：滑动窗口经典应用
public int lengthOfLongestSubstring(String s) {
Map<Character, Integer> map = new HashMap<>();
int maxLen = 0;
for (int left = 0, right = 0; right < s.length(); right++) {
char c = s.charAt(right);
if (map.containsKey(c)) {
left = Math.max(left, map.get(c) + 1); // 跳过重复字符
}
map.put(c, right);
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}5. 最长回文子串
题目：中心扩展法
public String longestPalindrome(String s) {
if (s == null || s.length() < 1) return "";
int start = 0, end = 0;
for (int i = 0; i < s.length(); i++) {
int len1 = expand(s, i, i); // 奇数长度
int len2 = expand(s, i, i + 1); // 偶数长度
int len = Math.max(len1, len2);
if (len > end - start) {
start = i - (len - 1) / 2;
end = i + len / 2;
}
}
return s.substring(start, end + 1);
}
private int expand(String s, int left, int right) {
while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
left--;
right++;
}
return right - left - 1;
}6. 字符串匹配算法（KMP）
题目：实现 strStr()
public int strStr(String haystack, String needle) {
if (needle.isEmpty()) return 0;
int[] next = getNext(needle);
int i = 0, j = 0;
while (i < haystack.length()) {
if (j == -1 || haystack.charAt(i) == needle.charAt(j)) {
i++;
j++;
} else {
j = next[j];
}
if (j == needle.length()) {
return i - j;
}
}
return -1;
}
private int[] getNext(String pattern) {
int[] next = new int[pattern.length()];
next[0] = -1;
int i = 0, j = -1;
while (i < pattern.length() - 1) {
if (j == -1 || pattern.charAt(i) == pattern.charAt(j)) {
i++;
j++;
next[i] = j;
} else {
j = next[j];
}
}
return next;
}7. 有效的括号
题目：栈的经典应用
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (char c : s.toCharArray()) {
if (c == '(') stack.push(')');
else if (c == '[') stack.push(']');
else if (c == '{') stack.push('}');
else if (stack.isEmpty() || stack.pop() != c)
return false;
}
return stack.isEmpty();
}8. 字母异位词分组
题目：HashMap的巧妙使用
public List<List<String>> groupAnagrams(String[] strs) {
Map<String, List<String>> map = new HashMap<>();
for (String s : strs) {
char[] chars = s.toCharArray();
Arrays.sort(chars);
String key = new String(chars);
map.computeIfAbsent(key, k -> new ArrayList<>()).add(s);
}
return new ArrayList<>(map.values());
}9. 编辑距离（动态规划）
题目：计算最小操作次数
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) dp[i][0] = i;
for (int j = 0; j <= n; j++) dp[0][j] = j;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = 1 + Math.min(dp[i - 1][j - 1],
Math.min(dp[i][j - 1], dp[i - 1][j]));
}
}
}
return dp[m][n];
}10. 字符串解码
题目：嵌套括号处理
public String decodeString(String s) {
Stack<Integer> numStack = new Stack<>();
Stack<StringBuilder> strStack = new Stack<>();
StringBuilder cur = new StringBuilder();
int num = 0;
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
num = num * 10 + (c - '0');
} else if (c == '[') {
numStack.push(num);
strStack.push(cur);
cur = new StringBuilder();
num = 0;
} else if (c == ']') {
StringBuilder tmp = cur;
cur = strStack.pop();
int repeat = numStack.pop();
for (int i = 0; i < repeat; i++) {
cur.append(tmp);
}
} else {
cur.append(c);
}
}
return cur.toString();
}