计算字符串中的特定字符（Java）
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count specific characters in a string (Java)
提问by user1432513
I have a homework assignment to count specific chars in string.
我有一个家庭作业来计算字符串中的特定字符。
For example: string = "America"
例如： string = "America"
The output should be = a appear 2 times, m appear 1 time, e appear 1 time, r appear 1 time, i appear 1 time and c appear 1 time
输出应该是 = a appear 2 times, m appear 1 time, e appear 1 time, r appear 1 time, i appear 1 time and c appear 1 time
public class switchbobo {
/**
* @param args
*/ // TODO Auto-generated method stub
public static void main(String[] args){
String s = "BUNANA";
String lower = s.toLowerCase();
char[] c = lower.toCharArray(); // converting to a char array
int freq =0, freq2 = 0,freq3 = 0,freq4=0,freq5 = 0;
for(int i = 0; i< c.length;i++) {
if(c[i]=='a') // looking for 'a' only
freq++;
if(c[i]=='b')
freq2++;
if (c[i]=='c') {
freq3++;
}
if (c[i]=='d') {
freq4++;
}
}
System.out.println("Total chars "+c.length);
if (freq > 0) {
System.out.println("Number of 'a' are "+freq);
}
}
}
code above is what I have done, but I think it is not make sense to have 26 variables (one for each letter). Do you guys have alternative result?
上面的代码是我所做的，但我认为有 26 个变量（每个字母一个）是没有意义的。你们有替代结果吗？
采纳答案by Hyman
Obviously your intuition of having a variable for each letter is correct.
显然，您对每个字母都有一个变量的直觉是正确的。
The problem is that you don't have any automated way to do the same work on different variables, you don't have any trivial syntax which helps you doing the same work (counting a single char frequency) for 26 different variables.
问题是你没有任何自动化的方法来对不同的变量做同样的工作，你没有任何简单的语法可以帮助你为 26 个不同的变量做同样的工作（计算单个字符频率）。
So what could you do? I'll hint you toward two solutions:
那你能做什么？我会向您提示两种解决方案：
- you can use an array (but you will have to find a way to map character
a-zto indices0-25, which is somehow trivial is you reason about ASCII encoding) - you can use a
HashMap<Character, Integer>which is an associative container that, in this situation, allows you to have numbers mapped to specific characters so it perfectly fits your needs
- 您可以使用数组（但您必须找到一种将字符映射
a-z到索引的方法0-25，这在某种程度上是微不足道的，因为您之所以使用 ASCII 编码） - 您可以使用 a
HashMap<Character, Integer>这是一个关联容器，在这种情况下，它允许您将数字映射到特定字符，因此它完全符合您的需求
回答by Sleiman Jneidi
You can use HashMapof Character key and Integer value.
您可以使用HashMap字符键和整数值。
HashMap<Character,Integer>
iterate through the string
遍历字符串
-if the character exists in the map get the Integer value and increment it.
-if not then insert it to map and set the integer value for 0
This is a pseudo code and you have to try coding it
这是一个伪代码，您必须尝试对其进行编码
回答by user1813084
I am using a HashMap for the solution.
我正在使用 HashMap 作为解决方案。
import java.util.*;
public class Sample2 {
/**
* @param args
*/
public static void main(String[] args)
{
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
String test = "BUNANA";
char[] chars = test.toCharArray();
for(int i=0; i<chars.length;i++)
{
if(!map.containsKey(chars[i]))
{
map.put(chars[i], 1);
}
map.put(chars[i], map.get(chars[i])+1);
}
System.out.println(map.toString());
}
}
Produced Output - {U=2, A=3, B=2, N=3}
生产输出 - {U=2, A=3, B=2, N=3}
回答by selva
int a[]=new int[26];//default with count as 0
for each chars at string
if (String having uppercase)
a[chars-'A' ]++
if lowercase
then a[chars-'a']++
回答by testUser
public class TestCharCount {
public static void main(String args[]) {
String s = "america";
int len = s.length();
char[] c = s.toCharArray();
int ct = 0;
for (int i = 0; i < len; i++) {
ct = 1;
for (int j = i + 1; j < len; j++) {
if (c[i] == ' ')
break;
if (c[i] == c[j]) {
ct++;
c[j] = ' ';
}
}
if (c[i] != ' ')
System.out.println("number of occurance(s) of " + c[i] + ":"
+ ct);
}
}
}
回答by user3326645
maybe you can use this
也许你可以用这个
public static int CountInstanceOfChar(String text, char character ) {
char[] listOfChars = text.toCharArray();
int total = 0 ;
for(int charIndex = 0 ; charIndex < listOfChars.length ; charIndex++)
if(listOfChars[charIndex] == character)
total++;
return total;
}
for example:
例如：
String text = "america";
char charToFind = 'a';
System.out.println(charToFind +" appear " + CountInstanceOfChar(text,charToFind) +" times");
回答by Gajender Singh
Count char 'l' in the string.
计算字符串中的字符“l”。
String test = "Hello";
int count=0;
for(int i=0;i<test.length();i++){
if(test.charAt(i)== 'l'){
count++;
}
}
or
或者
int count= StringUtils.countMatches("Hello", "l");
回答by Surender Thakran
In continuation to Hyman's answer the following code could be your solution. It uses the an array to store the frequency of characters.
继续Hyman的回答，以下代码可能是您的解决方案。它使用一个数组来存储字符的频率。
public class SwitchBobo
{
public static void main(String[] args)
{
String s = "BUNANA";
String lower = s.toLowerCase();
char[] c = lower.toCharArray();
int[] freq = new int[26];
for(int i = 0; i< c.length;i++)
{
if(c[i] <= 122)
{
if(c[i] >= 97)
{
freq[(c[i]-97)]++;
}
}
}
System.out.println("Total chars " + c.length);
for(int i = 0; i < 26; i++)
{
if(freq[i] != 0)
System.out.println(((char)(i+97)) + "\t" + freq[i]);
}
}
}
It will give the following output:
它将给出以下输出：
Total chars 6
a 2
b 1
n 2
u 1
