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Commit 49a3e8d

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feat: add basic knowledge
1 parent 90c2646 commit 49a3e8d

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7 files changed

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‎basic/README.md‎

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# 算法
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## 常见的排序算法
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- 冒泡排序
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- 插入排序

‎basic/sort/BubbleSort.java‎

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public class BubbleSort {
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private static void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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}
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public static void main(String[] args) {
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int[] nums = {1, 2, 7, 9, 5, 8};
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boolean hasChange = true;
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for (int i = 0; i < nums.length - 1 && hasChange; ++i) {
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hasChange = false;
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for (int j = 0; j < nums.length - 1 - i; ++j) {
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if (nums[j] > nums[j + 1]) {
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swap(nums, j, j + 1);
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hasChange = true;
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}
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}
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}
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}
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}

‎basic/sort/InsertionSort.java‎

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public class InsertionSort {
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public static void main(String[] args) {
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int[] nums = { 1, 2, 7, 9, 5, 8 };
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for (int i = 1, j, current; i < nums.length; ++i) {
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current = nums[i];
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for (j = i - 1; j >=0 && nums[j] > current; --j) {
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nums[j + 1] = nums[j];
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}
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nums[j + 1] = current;
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}
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}
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}

‎basic/sort/README.md‎

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# 常见的排序算法
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## 冒泡排序
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定义一个布尔变量 `hasChange`,用来标记每轮是否进行了交换。在每轮遍历开始时,将 `hasChange` 设置为 false。
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### 代码示例
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```java
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for (int i = 0; i < nums.length - 1; ++i) {
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for (int j = 0; j < nums.length - 1 - i; ++j) {
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if (nums[j] > nums[j + 1]) {
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swap(nums, j, j + 1);
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}
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}
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}
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```
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### 算法分析
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空间复杂度 O(1)、时间复杂度 O(n^2)。
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分情况讨论:
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1. 给定的数组按照顺序已经排好:只需要进行 `n-1` 次比较,两两交换次数为 0,时间复杂度为 O(n),这是最好的情况。
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2. 给定的数组按照逆序排列:需要进行 `n*(n-1)/2` 次比较,时间复杂度为 O(n^2),这是最坏的情况。
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3. 给定的数组杂乱无章。在这种情况下,平均时间复杂度 O(n^2)。
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因此,时间复杂度是 O(n^2),这是一种稳定的排序算法。
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> 稳定是指,两个相等的数,在排序过后,相对位置保持不变。
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## 插入排序
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与冒泡排序对比:
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- 在冒泡排序中,经过每一轮的排序处理后,数组后端的数是排好序的。
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- 在插入排序中,经过每一轮的排序处理后,数组前端的数是拍好序的。
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插入排序算法的基本思想是:不断将尚未排好序的数插入到已经排好序的部分。
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### 代码示例
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```java
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for (int i = 1, j, current; i < nums.length; ++i) {
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current = nums[i];
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for (j = i - 1; j >=0 && nums[j] > current; --j) {
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nums[j + 1] = nums[j];
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}
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nums[j + 1] = current;
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}
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```
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### 算法分析
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空间复杂度 O(1),时间复杂度 O(n^2)。
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分情况讨论:
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1. 给定的数组按照顺序排好序:只需要进行 n-1 次比较,两两交换次数为 0,时间复杂度为 O(n),这是最好的情况。
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2. 给定的数组按照逆序排列:需要进行 `n*(n-1)/2` 次比较,时间复杂度为 O(n^2),这是最坏的情况。
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3. 给定的数组杂乱无章:在这种情况下,平均时间复杂度是 O(n^2)。
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因此,时间复杂度是 O(n^2),这也是一种稳定的排序算法。
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class Solution {
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/**
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* 贪心算法
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*/
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public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {
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// 首先检查是否存在所有项目都可投资且初始资本 W >= max(Capital) 的情况。如果是,返回利润中前 k 个最大元素的和。
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boolean speedUp = true;
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for (int c : Capital) if (W < c) speedUp = false;
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if (speedUp) {
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PriorityQueue<Integer> heap = new PriorityQueue<>();
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for (int p : Profits) {
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heap.add(p);
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if (heap.size() > k) heap.poll();
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}
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for (int h : heap) W += h;
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return W;
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}
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int idx;
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int n = Profits.length;
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for (int i = 0; i < Math.min(k, n); ++i) {
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idx = -1;
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// 找到获取利润最多的项目
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for (int j = 0; j < n; ++j) {
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if (W >= Capital[j]) {
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if (idx == -1) idx = j;
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else if (Profits[idx] < Profits[j]) idx = j;
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}
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}
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// 找不到合适的项目
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if (idx == -1) break;
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// 累计当前项目的利润到总利润中,并把当前项目标记为"不可用"(设置成最大整数)
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W += Profits[idx];
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Capital[idx] = Integer.MAX_VALUE;
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}
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return W;
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}
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}
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class H2O {
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private Semaphore h = new Semaphore(2);
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private Semaphore o = new Semaphore(0);
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public H2O() {
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}
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public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
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// releaseHydrogen.run() outputs "H". Do not change or remove this line.
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h.acquire();
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releaseHydrogen.run();
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o.release();
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}
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public void oxygen(Runnable releaseOxygen) throws InterruptedException {
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// releaseOxygen.run() outputs "O". Do not change or remove this line.
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o.acquire(2);
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releaseOxygen.run();
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h.release(2);
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}
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}
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class FizzBuzz {
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private int n;
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public FizzBuzz(int n) {
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this.n = n;
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}
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private Semaphore fSema = new Semaphore(0);
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private Semaphore bSema = new Semaphore(0);
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private Semaphore fbSema = new Semaphore(0);
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private Semaphore nSema = new Semaphore(1);
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// printFizz.run() outputs "fizz".
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public void fizz(Runnable printFizz) throws InterruptedException {
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for (int i = 3; i <= n; i = i + 3) {
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if (i % 5 != 0) {
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fSema.acquire();
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printFizz.run();
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nSema.release();
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}
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}
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}
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// printBuzz.run() outputs "buzz".
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public void buzz(Runnable printBuzz) throws InterruptedException {
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for (int i = 5; i <= n; i = i + 5) {
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if (i % 3 != 0) {
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bSema.acquire();
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printBuzz.run();
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nSema.release();
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}
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}
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}
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// printFizzBuzz.run() outputs "fizzbuzz".
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public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
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for (int i = 15; i <= n; i = i + 15) {
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fbSema.acquire();
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printFizzBuzz.run();
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nSema.release();
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}
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}
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// printNumber.accept(x) outputs "x", where x is an integer.
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public void number(IntConsumer printNumber) throws InterruptedException {
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for (int i = 1; i <= n; i++) {
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nSema.acquire();
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if (i % 3 == 0 && i % 5 == 0) {
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fbSema.release();
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} else if (i % 3 == 0) {
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fSema.release();
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} else if (i % 5 == 0) {
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bSema.release();
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} else {
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printNumber.accept(i);
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nSema.release();
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}
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}
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}
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}

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