我的代码接受一个字符串，计算字母表中每个字符在字符串中出现的次数，然后显示数组的直方图，其中包含该字符在字符串中出现的次数的百分比信息。
我有问题，我的代码不显示正确的百分比，每个字符发生在字符串中，它是由几个百分点关闭。
例如，如果我输入"hello world"，L应该占字符串的30%，对吗?但我的代码说它占了字符串的27%我不太确定这是代码问题还是计算问题
public class LetterCounter {
private static int[] alphabetArray;
private static char ch;
double stringLength;
/**
* Creates an array to hold the total number of occurences for each character of the alphabet
*/
public LetterCounter()
{
alphabetArray = new int[26];
}
/**
* The method will go through the user inputted string, incrementing characters one by one
* when it comes across them in the string.
* @param input: the string that the user wants
*/
public void countLetters(String input) {
String userInput= input;
String s2 = userInput.toLowerCase();
for ( int i = 0; i < s2.length(); i++ ) {
char ch= s2.charAt(i);
if (ch >= 97 && ch <= 122){
alphabetArray[ch-'a']++;
}
}
stringLength = s2.length();
}
/**
* Calculates how many characters there are in the inputted string. Multiple strings that are input by the user will be
* added to the total character count.
* @return: the sum of how many characters there are in each string
*/
public int getTotalCount() {
int sum=0;
for (int i = 0; i < alphabetArray.length; i++) {
if(alphabetArray[i]>=0){
sum = 0;
char ch = (char) (i+97);
for(int total : alphabetArray) {
sum += total;
}
}
}
return sum;
}
/**
* This method will reset the character count to zero for every character.
*/
public void reset() {
for (int i : alphabetArray) {
if(alphabetArray[i]>=0){
alphabetArray[i]=0;
char ch = (char) (i+97);
System.out.println(ch +" : "+alphabetArray[i]);
}
}
}
/**
* The method prints out a histogram of the entire array, displaying the most commonly occuring letter as having 60 #s,
* with the rest of the letter's #s to the 60 #s. The percentage occurence of each character in the string is also displayed beside the character.
* @return: the histogram of the array
*/
public String toString() {
int max = alphabetArray[0];
int markCounter = 0;
double percent = 0.0;
String s = "";
for(int i =0; i<alphabetArray.length; i++) {
//finds the largest number of occurences for any letter in the string
if(alphabetArray[i] > max) {
max = alphabetArray[i];
}
}
for (int i = 0; i < alphabetArray.length; i++) {
//percent = (alphabetArray[i] /stringLength) * 100;
percent = Math.round((alphabetArray[i] /stringLength ) * 100);
markCounter = (alphabetArray[i] * 60) / max;
if(alphabetArray[i]>=0){
char ch = (char) (i+97);
System.out.print(ch +" : ");
}
for (int hash =0; hash <markCounter; hash++) {
System.out.print("#");
}
if(alphabetArray[i] >= 0){
System.out.print(" " + percent + "%"); //+ percent);
}
System.out.println();
}
return s;
}
}
您正在使用string.length()函数来计算字符串的长度。这个函数也会计算字符串长度中的空格。我猜你是在试图计算字符串中可用的总字符的百分比。
需要计算循环中的字符串长度，检查每个字符出现的频率。
public void countLetters(String input) {
String userInput= input;
String s2 = userInput.toLowerCase();
for ( int i = 0; i < s2.length(); i++ ) {
char ch= s2.charAt(i);
if (ch >= 97 && ch <= 122){
alphabetArray[ch-'a']++;
stringLength++;
}
}
}
这样您将获得字符串中字符的总数。
另外，您可能还想确定是否要包含字符而不是字母。特殊字符也可以是字符串的一部分。根据您的需求，字符串长度计数器将更新
String input = EditTextinput.getText().toString();输入=输入。
有时您可能只希望删除字符串开头或结尾的空格(而不是中间的空格)。如果是这种情况，你可以使用trim:
input = input.trim();