题目要求：（1）输出某个英文文本文件中 26 字母出现的频率，由高到低排列，并显示字母出现的百分比，精确到小数点后面两位
（2）输出单个文件中的前 N 个最常出现的英语单词
设计思想：
出现的问题：数据量太大，一维数组长度有限，对数据的处理过于复杂，时间复杂度太高
可能的解决方案（多选）：1、分批读入，操作
2、动态分配
源代码：
package main; import java.io.BufferedReader; import java.io.File; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.*; public class statist { public static void main(String[] args) throws Exception { mathmaticwords(); } public static void mathmaticnumber() { try { char shu[] = new char[1000000]; char zimu[] = new char[52]; int j = 0; int count[] = new int[52]; String pathname = "src/newAnalysis.txt"; File filename = new File(pathname); InputStreamReader reader = new InputStreamReader(new FileInputStream(filename)); BufferedReader br = new BufferedReader(reader); String line[] = new String[100]; for (int i = 0; i < line.length; i++) { line[i] = br.readLine(); System.out.println(line[i]); } br.close(); int k = 0; while (line[k] != null) { for (int i = 0; i < line[k].length(); i++) { shu[j] = line[k].charAt(i); j++; } k++; } // 匹配表 for (int i = 0; i < shu.length; i++) { switch (shu[i]) { case 'a': zimu[0] = 'a'; count[0]++; break; case 'b': zimu[1] = 'b'; count[1]++; break; case 'c': zimu[2] = 'c'; count[2]++; break; case 'd': zimu[3] = 'd'; count[3]++; break; case 'e': zimu[4] = 'e'; count[4]++; break; case 'f': zimu[5] = 'f'; count[5]++; break; case 'g': zimu[6] = 'g'; count[6]++; break; case 'h': zimu[7] = 'h'; count[7]++; break; case 'i': zimu[8] = 'i'; count[8]++; break; case 'j': zimu[9] = 'j'; count[9]++; break; case 'k': zimu[10] = 'k'; count[10]++; break; case 'l': zimu[11] = 'l'; count[11]++; break; case 'm': zimu[12] = 'm'; count[12]++; break; case 'n': zimu[13] = 'n'; count[13]++; break; case 'o': zimu[14] = 'o'; count[14]++; break; case 'p': zimu[15] = 'p'; count[15]++; break; case 'q': zimu[16] = 'q'; count[16]++; break; case 'r': zimu[17] = 'r'; count[17]++; break; case 's': zimu[18] = 's'; count[18]++; break; case 't': zimu[19] = 't'; count[19]++; break; case 'u': zimu[20] = 'u'; count[20]++; break; case 'v': zimu[21] = 'v'; count[21]++; break; case 'w': zimu[22] = 'w'; count[22]++; break; case 'x': zimu[23] = 'x'; count[23]++; break; case 'y': zimu[24] = 'y'; count[24]++; break; case 'z': zimu[25] = 'z'; count[25]++; break; case 'A': zimu[26] = 'A'; count[26]++; break; case 'B': zimu[27] = 'B'; count[27]++; break; case 'C': zimu[28] = 'C'; count[28]++; break; case 'D': zimu[29] = 'D'; count[29]++; break; case 'E': zimu[30] = 'E'; count[30]++; break; case 'F': zimu[31] = 'F'; count[31]++; break; case 'G': zimu[32] = 'G'; count[32]++; break; case 'H': zimu[33] = 'H'; count[33]++; break; case 'I': zimu[34] = 'I'; count[34]++; break; case 'J': zimu[35] = 'G'; count[35]++; break; case 'K': zimu[36] = 'K'; count[36]++; break; case 'L': zimu[37] = 'L'; count[37]++; break; case 'M': zimu[38] = 'M'; count[38]++; break; case 'N': zimu[39] = 'N'; count[39]++; break; case 'O': zimu[40] = 'O'; count[40]++; break; case 'P': zimu[41] = 'P'; count[41]++; break; case 'Q': zimu[42] = 'Q'; count[42]++; break; case 'R': zimu[43] = 'R'; count[43]++; break; case 'S': zimu[44] = 'S'; count[44]++; break; case 'T': zimu[45] = 'T'; count[45]++; break; case 'U': zimu[46] = 'U'; count[46]++; break; case 'V': zimu[47] = 'V'; count[47]++; break; case 'W': zimu[48] = 'W'; count[48]++; break; case 'X': zimu[49] = 'X'; count[49]++; break; case 'Y': zimu[50] = 'Y'; count[50]++; break; case 'Z': zimu[51] = 'Z'; count[51]++; } } int ci = 0; int sum = 0; System.out.println("短文中各字母出现情况统计如下："); for (int i = 0; i < 52; i++) { if (count[i] != 0) { ci++; sum += count[i]; } } ci = 0; for (int i = 0; i < 52; i++) { if (count[i] != 0) { ci++; System.out.println(count[i]); double a = (double) ((Math.round(count[i] * 100) / 100.0) / sum) * 100; double b = (double) (Math.round(a * 100) / 100.0); System.out.println(ci + ".字母" + zimu[i] + "的出现次数是:" + b); } } System.out.println("字母共计：" + sum + "个"); } catch (Exception e) { e.printStackTrace(); } } public static void mathmaticwords() throws IOException { BufferedReader br = new BufferedReader(new FileReader("src/newAnalysis.txt")); StringBuffer sb = new StringBuffer(); String text = null; while ((text = br.readLine()) != null) { sb.append(text);// 将读取出的字符追加到stringbuffer中 } br.close(); // 关闭读入流 String str = sb.toString().toLowerCase(); // 将stringBuffer转为字符并转换为小写 String[] words = str.split("[^(a-zA-Z)]+"); // 非单词的字符来分割，得到所有单词 Map<String, Integer> map = new HashMap<String, Integer>(); for (String word : words) { if (map.get(word) == null) { // 若不存在说明是第一次，则加入到map,出现次数为1 map.put(word, 1); } else { map.put(word, map.get(word) + 1); // 若存在，次数累加1 } } // 排序 List<Map.Entry<String, Integer>> list = new ArrayList<Map.Entry<String, Integer>>(map.entrySet()); Comparator<Map.Entry<String, Integer>> comparator = new Comparator<Map.Entry<String, Integer>>() { public int compare(Map.Entry<String, Integer> left, Map.Entry<String, Integer> right) { return (left.getValue().compareTo(right.getValue())); } }; // 集合默认升序升序 Collections.sort(list, comparator); for (int i = 0; i < list.size(); i++) {// 由高到低输出 System.out.println(list.get(list.size() - i - 1).getKey() + ":" + list.get(list.size() - i - 1).getValue()); } } }
结果截图：
总结：
通过这种小练习，认识到两点：1、无论程序板块或是软件项目多大，它都是由一个一个极小极小的基本程序单元组成，我们平常在写代码时要始终贯穿模块化基本化编程思想，把问题的规模和复杂性降维。
2、打铁还需自身硬，一段时间没有编写过小程序模块，竟然在面对问题时不知所措，这看出的我的基本功不是很扎实，计划复习JAVA基本语法，以期达到熟练。