问题
1.问题:以下字符串生成器的初始容量是多少?
StringBuilder sb = new StringBuilder("Able was I ere I saw Elba.");
答案:容量是初始字符串的长度+ 16:26 + 16 = 42。
2.考虑以下字符串:
String hannah = "Did Hannah see bees? Hannah did.";
a.问题:表达式hannah.length()显示的值是多少?
答: 32。
b.问题:方法调用hannah.charAt(12)返回的值是多少?
答: e。
c.问题:编写一个表达式,该表达式引用由hannah所引用的字符串中的字母b。
答: hannah.charAt(15)。
3.问题:以下表达式返回的字符串多长?字符串是什么?
"Was it a car or a cat I saw?".substring(9, 12)
答:长度为3个字符:car。它不包括car后的空间。
4.问题:在下面的ComputeResult程序中,执行每条带编号的行后,result的值是多少?
public class ComputeResult {public static void main(String[] args) {String original = "software";StringBuilder result = new StringBuilder("hi");int index = original.indexOf('a');/*1*/ result.setCharAt(0, original.charAt(0));/*2*/ result.setCharAt(1, original.charAt(original.length()-1));/*3*/ result.insert(1, original.charAt(4));/*4*/ result.append(original.substring(1,4));/*5*/ result.insert(3, (original.substring(index, index+2) + " "));System.out.println(result);}}
回答:
siseswesweoftswear oft
练习题
练习:展示两种将以下两个字符"Hi, mom."串连接在一起以得到字符串的方法:
String hi = "Hi, ";String mom = "mom.";
答: hi.concat(mom)和hi + mom。
练习:编写一个程序,使用您的全名来计算您的姓名首字母并将其显示出来。
回答: ComputeInitials
public class ComputeInitials {public static void main(String[] args) {String myName = "Fred F. Flintstone";StringBuffer myInitials = new StringBuffer();int length = myName.length();for (int i = 0; i < length; i++) {if (Character.isUpperCase(myName.charAt(i))) {myInitials.append(myName.charAt(i));}}System.out.println("My initials are: " + myInitials);}}
练习:字谜是通过一个单词或短语的字母换位而得到的另一个单词或短语;例如,“parliament”是“partial men”的字谜,“software”是“swear oft”的字谜。编写一个程序,弄清楚一个字符串是否是另一个字符串的字谜。该程序应忽略空格和标点符号。
回答: Anagram
public class Anagram {public static boolean areAnagrams(String string1,String string2) {String workingCopy1 = removeJunk(string1);String workingCopy2 = removeJunk(string2);workingCopy1 = workingCopy1.toLowerCase();workingCopy2 = workingCopy2.toLowerCase();workingCopy1 = sort(workingCopy1);workingCopy2 = sort(workingCopy2);return workingCopy1.equals(workingCopy2);}protected static String removeJunk(String string) {int i, len = string.length();StringBuilder dest = new StringBuilder(len);char c;for (i = (len - 1); i >= 0; i--) {c = string.charAt(i);if (Character.isLetter(c)) {dest.append(c);}}return dest.toString();}protected static String sort(String string) {char[] charArray = string.toCharArray();java.util.Arrays.sort(charArray);return new String(charArray);}public static void main(String[] args) {String string1 = "Cosmo and Laine:";String string2 = "Maid, clean soon!";System.out.println();System.out.println("Testing whether the following "+ "strings are anagrams:");System.out.println(" String 1: " + string1);System.out.println(" String 2: " + string2);System.out.println();if (areAnagrams(string1, string2)) {System.out.println("They ARE anagrams!");} else {System.out.println("They are NOT anagrams!");}System.out.println();}}
