以下是我的代码：
char[] array = new char[26] ;
int index = 0 ;
int letter = 0 ;
int countA = 0 ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
char[] count = new char[sentenceFinal.length()] ;
for (char c = 'a'; c <= 'z'; c++)
{
array[index++] = c ;
for(int i = 0; i < sentenceFinal.length(); i++)
{
if(sentenceFinal.charAt(i) == c)
count[letter++] = c ;
//if(sentenceFinal.charAt(i) == 'a')
//countA++ ;
}
}
String result = new String(count) ; // Convert to a string.
System.out.println("\n" + result) ;
System.out.println("\nTotal number of letters is " + result.length()) ;
System.out.println(countA) ;
}
else
{
System.out.println("You forgot a period. Try again.") ;
}
我很难计算给定句子中有多少个a，b，c等。我有一种方法可以做到，而这就是
//if(sentenceFinal.charAt(i) == 'a')
//countA++ ;
我可以一直创建到z。有没有更有效的方法？
注意：请勿使用Hashmap或任何其他高级技术。
答案 0 :(得分：1)
不需要消除空格。这只是您正在做的其他工作。
int countOfLetters = 0 ;
String sentence = "Once upon a time..." ;
sentence = sentence.toLowerCase();
int[] countOfAlphabets = new int[26];
for (int i = 0; i < sentence.length(); i++) {
if (sentence.charAt(i) >= 'a' && sentence.charAt(i) <= 'z') {
countOfAlphabets[sentence.charAt(i) - 97]++;
countOfLetters++;
}
}
因此，countOfLetters将为您提供字母总数。
例如，如果要计数，例如，您要计数'c'，
您可以通过访问countOfAlphabets之类的countOfAlphabets['c' - 97]数组（97是'a'的ASCII值）来获取它
答案 1 :(得分：0)
使用int数组letterCounts来存储每个字母的计数。假设字母的大小写可以忽略，则letterCounts数组的长度为26。
遍历字符串的字符并更新数组中的相应整数。使用其ASCII值查找相应的索引，如下所示。
letterCounts[c - 97]++
97是ASCII值'a'，其计数需要存储在索引0中。
通过这种方式，从字符的ASCII值中减去97将得到该字符的相应索引。
注意：这是假设您要存储小写字母的计数。
答案 2 :(得分：0)
很漂亮，没有使用映射，但这将计算字符串中的所有字符。
您可能需要修改以排除空格等内容。
public class Main {
public static void main(String[] args) {
String sentence = "Once upon a time...";
// Create an array of size 256 ASCII_SIZE
int count[] = new int[256];
int length = sentence.length();
// Initialize count array index
for (int i = 0; i < length; i++)
count[sentence.charAt(i)]++;
// Create an array of given String size
char chars[] = new char[sentence.length()];
for (int i = 0; i < length; i++) {
chars[i] = sentence.charAt(i);
int find = 0;
for (int j = 0; j <= i; j++) {
// If any matches found
if (sentence.charAt(i) == chars[j])
find++;
}
if (find == 1) {
System.out.println("Occurrence of " + sentence.charAt(i) + " is:" + count[sentence.charAt(i)]);
}
}
}
}
哪个输出：
Occurrence of O is:1
Occurrence of n is:2
Occurrence of c is:1
Occurrence of e is:2
Occurrence of is:3
Occurrence of u is:1
Occurrence of p is:1
Occurrence of o is:1
Occurrence of a is:1
Occurrence of t is:1
Occurrence of i is:1
Occurrence of m is:1
Occurrence of . is:3
答案 3 :(得分：0)
检查以下代码您可以拥有26个长度的数组，索引将根据字母的存在而增加。
public void getResult(){
int [] charCount = new int [26];
int countA = 0 ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
char[] sentenceCharArray = sentenceFinal.toCharArray();
//char a = 97;
for (int i = 0; i <sentenceCharArray.length ; i++) {
int index = sentenceCharArray[i] - 97 ;
if(index >= 0 && index <= 26) {
charCount[index] += 1;
}
}
System.out.print("Result : ");
for (int i = 0; i < charCount.length ; i++) {
System.out.print(charCount [i]+" , ");
}
System.out.println("\nTotal number of letters is " + sentenceCharArray.length) ;
}
else
{
System.out.println("You forgot a period. Try again.") ;
}
}
答案 4 :(得分：0)
由于美国字母表中有26个字母，因此您可以使用大小为26的int[]
int[] letterCount = new int[26];
用于保存每个字母的计数，其中索引0代表“ a”，索引1代表“ b”，等等...
遍历句子时，请检查您所使用的字符是否为字母Character.isLetter()，然后递增表示字母的数组中的元素。
letterCount[letter - 'a']++;
我们从字母中减去'a'以给我们正确的索引。
package stackoverflow;
public class Question {
public static void main(String[] args) {
String sentence = "The quick brown fox jumps over the lazy dog.";
int[] letterCount = new int[26];
if (sentence.contains(".")) {
// toLowerCase() the sentence since we count upper and lowercase as the same
for (char letter : sentence.toLowerCase().toCharArray()) {
if (Character.isLetter(letter)) {
letterCount[letter - 'a']++;
}
}
// Display the count of each letter that was found
int sumOfLetters = 0;
for (int i = 0; i < letterCount.length; i++) {
int count = letterCount[i];
if (count > 0) {
System.out.println((char)(i + 'a') + " occurs " + count + " times");
sumOfLetters += count;
}
}
System.out.println("Total number of letters is " + sumOfLetters);
} else {
System.out.println("You forgot a period. Try again.");
}
}
}
a occurs 1 times
b occurs 1 times
c occurs 1 times
d occurs 1 times
e occurs 3 times
f occurs 1 times
g occurs 1 times
h occurs 2 times
i occurs 1 times
j occurs 1 times
k occurs 1 times
l occurs 1 times
m occurs 1 times
n occurs 1 times
o occurs 4 times
p occurs 1 times
q occurs 1 times
r occurs 2 times
s occurs 1 times
t occurs 2 times
u occurs 2 times
v occurs 1 times
w occurs 1 times
x occurs 1 times
y occurs 1 times
z occurs 1 times
Total number of letters is 35
使用Java 8和使用chars()中的String有什么问题？有了它，您可以用更少的代码完成相同的事情。对于字母总数，我们只使用String.replaceAll()并用模式String从[^A-Za-z]除去所有非字母，然后使用结果的length()。
package stackoverflow;
import java.util.function.Function;
import java.util.stream.Collectors;
public class Question {
public static void main(String[] args) {
String sentence = "The quick brown fox jumps over the lazy dog.";
System.out.println(sentence.toLowerCase().chars()
// Change the IntStream to a stream of Characters
.mapToObj(c -> (char)c)
// Filter out non lower case letters
.filter(c -> 'a' <= c && c <= 'z')
// Collect up the letters and count them
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting())));
System.out.println("Total letter count is " + sentence.replaceAll("[^A-Za-z]", "").length());
}
}
{a=1, b=1, c=1, d=1, e=3, f=1, g=1, h=2, i=1, j=1, k=1, l=1, m=1, n=1, o=4, p=1, q=1, r=2, s=1, t=2, u=2, v=1, w=1, x=1, y=1, z=1}
Total letter count is 35
答案 5 :(得分：0)
如果Regex不会被视为高科技
想法很简单：删除所有字母并从原始字符串长度中减去输出以获得计数器
String sentence = "Once upon a time...";
String noLetterString = sentence.replaceAll("[a-zA-Z]", "");
int counterLetter = sentence.length() - noLetterString.length();
System.out.println("counter:" + counterLetter);
通过老式编程 这里的想法是相反的，只添加字母
String sentence = "Once upon a time...";
String lowerCase = sentence.toLowerCase(); // to avoid comparison to UpperCase letters
StringBuilder counterStr = new StringBuilder();
for (char l : lowerCase.toCharArray()) {
if (l >= 'a' && l <= 'z') {
counterStr.append(l);
}
}
System.out.println("counterStr:" + counterStr);
System.out.println("counter:" + counterStr.length());
答案 6 :(得分：-1)
这是更新代码：
int[] array = new int[26] ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
for (char c : sentenceFinal.toCharArray())
{
System.out.println(c+" "+(c-97));
array[c-97] += 1;
}
// System.out.println("\n" + Arrays.toString(array)) ;
for(int i=0; i< array.length;i++) {
if(array[i] != 0) {
char c = (char)(i+97);
System.out.println(c+" occured "+ array[i]+" times");
}
}
}
else
{
System.out.println("You forgot a period. Try again.") ;
}