递归分治法在快速排序中的应用 java以界面的方式实现

递归分治法在快速排序中的应用

 

分治法的基本思想

• §分治法的基本思想是将一个规模为n的问题分解为k个规模较小的子问题，这些子问题互相独立且与原问题相同。 
•         §对这k个子问题分别求解。如果子问题的规模仍然不够小，则再划分为k个子问题，如此递归的进行下去，直到问题规模足够小，很容易求出其解为止。 
•         §将求出的小规模的问题的解合并为一个更大规模的问题的解，自底向上逐步求出原来问题的解。 
•         §分治法的设计思想是，将一个难以直接解决的大问题，分割成一些规模较小的相同问题，以便各个击破，分而治之。

      Ø由分治法产生的子问题往往是原问题的较小模式，这就为使用递归技术提供了方便。在这种情况下，反复应用分治手段，可以使子问题与原问题类型一致而其规模却不断缩小，最终使子问题缩小到很容易直接求出其解。这自然导致递归过程的产生。

      Ø分治与递归像一对孪生兄弟，经常同时应用在算法设计之中，并由此产生许多高效算法。

分治法的适用条件

§分治法所能解决的问题一般具有以下几个特征：

•该问题的规模缩小到一定的程度就可以容易地解决；

•该问题可以分解为若干个规模较小的相同问题，即该问题具有最优子结构性质

•利用该问题分解出的子问题的解可以合并为该问题的解；

•该问题所分解出的各个子问题是相互独立的，即子问题之间不包含公共的子问题。

§这条特征涉及到分治法的效率，如果各子问题是不独立的，则分治法要做许多不必要的工作，重复地解公共的子问题，此时虽然也可用分治法，但一般用动态规划较好。

public class mainRecursion {/** * @param 主类 */public static void main(String[] args) {// TODO Auto-generated method stub new recursionFrame();}} 

import java.util.*;public class numberTaxis {public boolean isNumber(String number) {// 检验数字是否合理StringTokenizer tokenizer = new StringTokenizer(number, ",");boolean bool = true;while (tokenizer.hasMoreTokens()) {try {Double.valueOf(tokenizer.nextToken());} catch (Exception e) {bool = false;}}return bool;}double sumNumber[];public String uptaxis(String number, boolean updown) {// 排序int i = 0;String s = "";StringTokenizer tokenizer = new StringTokenizer(number, ",");sumNumber = new double[tokenizer.countTokens()];//字符串的解析while (tokenizer.hasMoreTokens()) {String d = tokenizer.nextToken();sumNumber[i] = Double.valueOf(d);//将字符串转换为双字节类型i++;}QSort(sumNumber, 0, sumNumber.length - 1);// 快速排序（升序）// 如果选择降序if (!updown) {double num;int low = 0, high = sumNumber.length - 1;for (; low < high; low++, high--) {num = sumNumber[low];//数组交换位子sumNumber[low] = sumNumber[high];sumNumber[high] = num;}}for (i = 0; i < sumNumber.length - 1; i++) {s += String.valueOf(sumNumber[i]) + "，";}return s + String.valueOf(sumNumber[sumNumber.length - 1]);} //搜索字符串public String search(String s, boolean updown) {String str = "";int a = searchNumber(Double.valueOf(s), updown);// 调用折半查找if (a != -1)str = "查找成功，位序为：" + String.valueOf(a + 1);elsestr = "没有找到该元素！";return str;}public int searchNumber(double num, boolean updown) {// 折半查找算法int left = 0, right = sumNumber.length - 1, middle = -1;while (left <= right) {middle = (left + right) / 2;if (num == sumNumber[middle])return middle;if (updown) {//如果数组是升序if (num > sumNumber[middle])left = middle + 1;elseright = middle - 1;} else {//数组是降序if (num > sumNumber[middle])right = middle - 1;elseleft = middle + 1;}}return -1;}// 快速排序算法public int partition(double num[], int low, int high) {double m = num[low];// 记录关键字while (low < high) {while (low < high && num[high] >= m)--high;// 将比记录小的记录移到低端num[low] = num[high];while (low < high && num[low] <= m)++low;// 将比记录大的记录移到高端num[high] = num[low];}num[low] = m;// 返回记录轴的位置return low;} //快速排序递归 的实现public void QSort(double num[], int low, int high) {if (low < high) {// 长度大于一int pivotloc = partition(num, low, high);// 将数据一分为二QSort(num, low, pivotloc - 1);// 递归排序QSort(num, pivotloc + 1, high);}}public void Quick(double num[], boolean updown) {// 选择排序算法int i = 0, j, k;double n;for (i = 0; i < sumNumber.length; i++) {k = i;for (j = i + 1; j < sumNumber.length; j++) {if (updown) {//如果数组是升序if (sumNumber[j] < sumNumber[k])k = j;} else {if (sumNumber[j] > sumNumber[k])k = j;}}n = sumNumber[k];sumNumber[k] = sumNumber[i];sumNumber[i] = n;}}} 

import java.awt.Dimension;import java.awt.Toolkit;import java.awt.event.ActionEvent;import java.awt.event.ActionListener;import java.awt.event.WindowAdapter;import java.awt.event.WindowEvent;import javax.swing.*;public class recursionFrame extends JFrame implements ActionListener {private JLabel firstNumberLabel, secondNumberLabel, searchLabel,resultLabel, result;private JButton upButton, downButton;private JTextField firstNumber, secondNumber, searchNumber;numberTaxis numbertaxis;recursionFrame() {super("递归分治法应用");numbertaxis = new numberTaxis();mainFrame();}boolean updown = true;public void actionPerformed(ActionEvent e) {// TODO 事件的处理if (e.getSource() == firstNumber) {action(true);} else if (e.getSource() == downButton) {action(false);updown = false;} else if (e.getSource() == upButton) {action(true);updown = true;} else if (e.getSource() == searchNumber) {String str1 = searchNumber.getText();String str2 = secondNumber.getText();boolean isnumber = numbertaxis.isNumber(searchNumber.getText());if (!str2.equals("") && !str1.equals("") && isnumber) {result.setText(numbertaxis.search(str1, updown));} else {JOptionPane.showMessageDialog(this, "您输入的搜索数据有错误请从新输入！","警告对话框", JOptionPane.WARNING_MESSAGE);searchNumber.setText("");}}}public void mainFrame() {firstNumberLabel = new JLabel("请输入一组数据，以逗号隔开(默认为升序排列)：");secondNumberLabel = new JLabel("排序结果：");searchLabel = new JLabel("请输入要查询的数据，以回车键确定:");resultLabel = new JLabel("查询结果是：");result = new JLabel(" ");upButton = new JButton("升序排列");downButton = new JButton("降序排列");firstNumber = new JTextField(50);secondNumber = new JTextField(50);searchNumber = new JTextField(10);upButton.addActionListener(this);downButton.addActionListener(this);firstNumber.addActionListener(this);searchNumber.addActionListener(this);this.add(firstNumberLabel);firstNumberLabel.setBounds(10, 10, 290, 25);this.add(firstNumber);firstNumber.setBounds(10, 35, 270, 25);this.add(upButton);upButton.setBounds(20, 65, 100, 25);this.add(downButton);downButton.setBounds(150, 65, 100, 25);this.add(secondNumberLabel);secondNumberLabel.setBounds(10, 95, 70, 25);this.add(secondNumber);secondNumber.setBounds(10, 120, 270, 25);this.add(searchLabel);searchLabel.setBounds(10, 150, 230, 25);this.add(searchNumber);searchNumber.setBounds(230, 150, 50, 25);this.add(resultLabel);resultLabel.setBounds(10, 180, 100, 30);this.add(result);result.setBounds(10, 205, 200, 30);Toolkit tool = getToolkit();// 获取屏幕的大小Dimension dim = tool.getScreenSize();this.setBounds(dim.width / 2 - 150, dim.height / 2 - 100, 300, 300);this.setLayout(null);this.setResizable(false);this.setVisible(true);this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);//this.addWindowListener(new WindowAdapter()//{//public void windowClosing(WindowEvent e){//System.exit(0);//}//}//);}public void action(boolean updown) {// 对输入的数据验证和排序if (numbertaxis.isNumber(firstNumber.getText())&& !firstNumber.getText().equals("")) {secondNumber.setText(numbertaxis.uptaxis(firstNumber.getText(),updown));} else {JOptionPane.showMessageDialog(this, "您输入的数据有错误请从新输入！", "警告对话框",JOptionPane.WARNING_MESSAGE);}}}