【每日一练】20231023

统计每个字符出现的次数相关问题 方法一：map的put方法+遍历 public class Test { public static void main(String[] args) { StringBuilder sb = new StringBuilder(""); Random ran = new Random(); for(int i=0;i<2000000;i++) { sb.append((char) ('a' + ran.nextInt(27))); } String str = sb.toString(); //System.out.println(sb); //统计一下每个字符出现的次数？出现最多？ Map<Character, Integer> map = new HashMap<>(); long s = System.currentTimeMillis(); for (int i = 0; i < str.length(); i++) { Character ch = str.charAt(i); if (Objects.isNull(map.get(ch))) { map.put(ch, 1); } else { map.put(ch, map.get(ch) + 1); } } long e = System.currentTimeMillis(); System.out.println((e-s)+"ms"); // System.out.println(map); } } 方法二：用数组 public class Test2 { public static void main(String[] args) { StringBuilder sb = new StringBuilder(""); //a-z Random ran = new Random(); for(int i=0;i<2000000;i++) { sb.append((char) ('a' + ran.nextInt(26))); } String str = sb.toString(); // System.out.println(str); int[] chs =new int[26]; //chs[0] chs[25] 0-a, 25-z long s = System.currentTimeMillis(); for(int i =0; i<str.length();i++){ char c = str.charAt(i); //if c is a chs[c-'a']+=1; } long e = System.currentTimeMillis(); System.out.println((e-s)+"ms"); // // for(int i=0;i<chs.length;i++){ // System.out.print((char)(i+'a')+","); // System.out.println(chs[i]); // } } } 方法三：map的getOrDefault方法 public class Test3 { public static void main(String[] args) { String str = "The so-called 'multifunctional' is not adding many elements, " + "but subtracting them. Only by leaving blank can ten thousand possibilities be created."; // 创建HashMap用于存储字母和出现次数 Map<Character, Integer> countMap = new HashMap<>(); // 遍历字符串的每个字符 for (char c : str.toCharArray()) { // 判断是否为字母 if (Character.isLetter(c)) { // 将字母转为小写形式 char lowercase = Character.toLowerCase(c); // 更新字母的出现次数，使用Map的getOrDefault方法来获取该字母已有的计数并加一 countMap.put(lowercase, countMap.getOrDefault(lowercase, 0) + 1); } } // 遍历Map输出每个字母及其出现次数 for (char c : countMap.keySet()) { int count = countMap.get(c); System.out.println(c + ": " + count); } } }

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