使用表格查找用户最常出现的字母[复制]

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Java program to find the character that appears the most number of times in a String? 8 answers

Java程序找到在String中出现次数最多的字符? 8个答案

I got a task from my university today:

我今天从大学得到了一个任务:

Write a program that reads a ( short ) text from the user and prints the so called max letter (most common character in string) , that the letter which the greatest number of occurrences of the given text . Here it is enough to look at English letters (A- Z) , and not differentiate between uppercase and lowercase letters in the count of the number of occurrences .

编写一个程序,从用户读取(短)文本并打印所谓的最大字母(字符串中最常见的字符),即给定文本出现次数最多的字母。这里足以查看英文字母(A-Z),而不是区分出现次数的大写和小写字母。

For example, if : text = " Ada bada " so should the print show the most common character, this example it would be a.

例如,如果:text =“Ada bada”,那么打印应该显示最常见的字符,这个例子就是a。

This is an introductory course, so in this submission we do not need to use the " scanner - class" . We have not gone through this so much. The program will use the show message input two get the text from user .

这是一个入门课程,因此在本次提交中我们不需要使用“scanner - 类”。我们没有经历过这么多。该程序将使用show message输入两个来自用户的文本。

Info: The program shall not use while loop ( true / false ) , "return " statement / "break " statement .

信息:程序不应使用while循环(true / false),“return”语句/“break”语句。

I've been struggling with how I can get char values into a table.. am I correct I need to use array to search for most common character? I think I need to use the binarySearch, but that only supports int not char.

我一直在努力学习如何将char值放入表中。我是否正确我需要使用数组来搜索最常见的字符?我想我需要使用binarySearch,但是它只支持int而不是char。

I'll be happy for any answers. hint's and solutions. etc.. if you're very kind a full working program, but again please don't use the things I have written down in the "info" section above.

我会很高兴得到任何答案。提示和解决方案。等等。如果你是一个非常善良的完整工作程序,但请再次请不要使用我在上面的“信息”部分写下的内容。

My code:

    String text = showInputDialog("Write a short text: "); 

    //format string to char

      String a = text;
      char c = a.charAt(4);
/*with this layout it collects number 4 character in the text and print out.
* I could as always go with many char c... but that wouldn't be a clean program * code.. I think I need to make it into a for-loop.. I have only worked with * *for-loops with numbers, not char (letters).. Help? :) 
*/

    out.print( text + "\n" + c)

//each letter into 1 char, into table
//search for most used letter

3 个解决方案

#1

Here's the common logic:

这是常见的逻辑:

split your string into chars

将你的字符串分成字符

loop over the chars

在字符上循环

store the occurrences in a hash, putting the letter as key and occurrences as value

将事件存储在哈希中,将字母作为键,将出现值作为值

return the highest value in the hash

返回哈希值中的最高值

As how to split string into chars, etc., you can use Google. :)

至于如何将字符串拆分为字符等,您可以使用Google。 :)

Here's a similar question.

这是一个类似的问题。

#2

There's a common program asked to write in schools to calculate the frequency of a letter in a given String. The only thing you gotta do here is find which letter has the maximum frequency. Here's a code that illustrates it:

有一个共同的程序被要求在学校写,以计算给定字符串中的字母的频率。你要做的唯一事情是找到哪个字母的最大频率。这是一个说明它的代码:

String s <--- value entered by user
char max_alpha=' '; int max_freq=0, ct=0;
char c;
for(int i=0;i<s.length();i++){
    c=s.charAt(i);
    if((c>='a'&&c<='z')||(c>='A'&&c<='Z')){
        for(int j=0;j<s.length();j++){
            if(s.charAt(j)==c)
            ct++;
        } //for j
    }
    if(ct>max_freq){
        max_freq=ct;
        max_alpha=c;
    }
    ct=0;
    s=s.replace(c,'*');
}
System.out.println("Letter appearing maximum times is "+max_alpha);
System.out.println(max_alpha+" appears "+max_freq+" times");

NOTE: This program presumes that all characters in the string are in the same case, i.e., uppercase or lowercase. You can convert the string to a particular case just after getting the input.

注意:此程序假定字符串中的所有字符都是相同的大小写,即大写或小写。获取输入后,您可以将字符串转换为特定的大小写。

#3

I guess this is not a good assigment, if you are unsure about how to start. I wish you for having better teachers!

如果你不确定如何开始,我想这不是一个好的分配。我希望你有更好的老师!

So you have a text, as:

所以你有一个文本,如:

String text = showInputDialog("Write a short text: ");

The next thing is to have a loop which goes trough each letter of this text, and gets each char of it:

接下来的事情是有一个循环,它通过这个文本的每个字母,并获取它的每个字符:

for (int i=0;i<text.length();i++) {
    char c=text.charAt(i);
}

Then comes the calculation. The easiest thing is to use a hashMap. I am unsure if this is a good topic for a beginners course, so I guess a more beginner friendly solution would be a better fit.

然后是计算。最简单的方法是使用hashMap。我不确定这对于初学者课程是否是一个好主题,所以我想一个更适合初学者的解决方案会更合适。

Make an array of integers - this is the "table" you are referring to. Each item in the array will correspond to the occurrance of one letter, e.g. histogram[0] will count how many "A", histogram[1] will count how many "B" you have found.

创建一个整数数组 - 这是你所指的“表”。数组中的每个项目将对应于一个字母的出现,例如直方图[0]将计算“A”的数量,直方图[1]将计算您找到的“B”的数量。

int[] histogram = new int[26]; // assume English alphabet only

for (int i=0;i<histogram.length;i++) {
    histogram[i]=0;
}

for (int i=0;i<text.length();i++) {
    char c=Character.toUppercase(text.charAt(i));

    if ((c>=65) && (c<=90)) {
        // it is a letter, histogram[0] contains occurrences of "A", etc.
        histogram[c-65]=histogram[c-65]+1;
    }
}

Then finally find the biggest occurrence with a for loop...

然后终于找到一个for循环最大的事件......

int candidate=0;
int max=0;
for (int i=0;i<histogram.length;i++) {
    if (histogram[i]>max) {
        // this has higher occurrence than our previous candidate
        max=histogram[i];
        candidate=i; // this is the index of char, i.e. 0 if A has the max occurrence
    }
}

And print the result:

并打印结果:

System.out.println(Character.toString((char)(candidate+65));

Note how messy this all comes as we use ASCII codes, and only letters... Not to mention that this solution does not work at all for non-English texts.

请注意这一切是多么混乱,因为我们使用ASCII码,只有字母......更不用说这个解决方案根本不适用于非英语文本。

If you have the power of generics and hashmaps, and know some more string functions, this mess can be simplified as:

如果您具有泛型和散列图的强大功能,并且知道更多的字符串函数,那么这个混乱可以简化为:

String text = showInputDialog("Write a short text: "); 
Map<Char,Integer> histogram=new HashMap<Char,Integer>();

for (int i=0;i<text.length();i++) {
    char c=text.toUppercase().charAt(i));
    if (histogram.containsKey(c)) {
        // we know this letter, increment its occurrence
        int occurrence=histogram.get(c);
        histogram.put(c,occurrence+1);
    }
    else { 
        // we dunno this letter yet, it is the first occurrence
        histogram.put(c,1);
    }
}

char candidate=' ';
int max=0;
for (Char c:histogram.keySet()) {
    if (histogram.get(c)>max) {
        // this has higher occurrence than our previous candidate
        max=histogram.get(c);
        candidate=c; // this is the char itself
    }
}

System.out.println(c);

small print: i didn't run this code but it shall be ok.

小字体:我没有运行此代码,但它应该没问题。

#1