Question

当我试图分别分析字符串的每个单词时，第一个单词做得很好，但其余的都没有成功。帮我纠正这个JAVA代码以获得正确的输出：

public class Test {

 public static void main(String[] args) {


            int count1=0;
    int chars_not_x=0 ;

String str = "xyz xyxzghz zyxzz";
    String[] words = str.split("\\s");

for(int m=0; m<words.length; m++){

            System.out.println(words[m]);

            System.out.println("Total characters in the word: "+words[m].length());
    for(int n=0; n<words[m].length(); n++){
if(words[m].charAt(n)=='x'){count1++;}}

           System.out.println("Number of x :"+count1);
           chars_not_x= words[m].length()- count1; 
           System.out.println("Chars other than x: "+chars_not_x);

           System.out.println("\n");


  }} }

代码的输出是＆＃34;

xyz
Total characters in the word: 3
Number of x :1
Chars other than x: 2


xyxzghz
Total characters in the word: 7
Number of x :3
Chars other than x: 4


zyxzz
Total characters in the word: 5
Number of x :4
Chars other than x: 1

所需的输出：

xyz
Total characters in the word: 3
Number of x :1
Chars other than x: 2


xyxzghz
Total characters in the word: 7
Number of x :2
Chars other than x: 5


zyxzz
Total characters in the word: 5
Number of x :1
Chars other than x: 4

Answer 1

根据需要使用工作代码 - ＆gt;

public class Test {
public static void main(String[] args) {

int count1=0;
int chars_not_x=0 ;

String str = "xyz xyxzghz zyxzz";
String[] words = str.split("\\s");

for(int m=0; m<words.length; m++){

count1 = 0;    // Add this

System.out.println(words[m]);

System.out.println("Total characters in the word: "+words[m].length());
for(int n=0; n<words[m].length(); n++){
if(words[m].charAt(n)=='x'){count1++;}}

       System.out.println("Number of x :"+count1);
       chars_not_x= words[m].length()- count1; 
       System.out.println("Chars other than x: "+chars_not_x);
       System.out.println("\n");


}} }

Answer 2

问题是您在错误的位置声明了递增的值，因此错误地使用了它们。我已经改变了你的代码，看看它是否对你有帮助：

public static void main(String[] args) {

    String str = "xyz xyxzghz zyxzz";
    String[] words = str.split(" ");

    for (String word : words) {

        int xFound = 0;
        int nonXFound = 0;
        String[] chars = word.split("(?!^)");
        for (String current : chars) {
            if ("x".equals(current)) {
                xFound++;
            } else {
                nonXFound++;
            }
        }

        System.out.println("Word [" + word + "] has [" + chars.length + "] characters, and contains [" + xFound + "] x and [" + nonXFound + "] non-x.");
    }
}

将字符串拆分为单词
将每个单词拆分为与x
单词的长度是分割单词的String数组的大小

输出：

Word [xyz]有[3]个字符，包含[1] x和[2] non-x。

Word [xyxzghz]有[7]个字符，包含[2] x和[5] non-x。

Word [zyxzz]有[5]个字符，包含[1] x和[4] non-x。

注意：用于拆分我使用的字符的正则表达式是"(?!^)"而不是""，因为后者生成一个不需要的空白字符作为数组中的第0个元素（试试看）。分割为字符或单字符字符串数组可以用不同的方式完成，这只是一个例子。

Answer 3

确保重置每个单词的x计数：

for(int m=0; m<words.length; m++){
    System.out.println(words[m]);
    System.out.println("Total characters in the word: "+words[m].length());
    for(int n=0; n<words[m].length(); n++){
        count1 =0; //**missing** 
        if(words[m].charAt(n)=='x'){count1++;}
    }

    System.out.println("Number of x :"+count1);
    chars_not_x= words[m].length()- count1; 
    System.out.println("Chars other than x: "+chars_not_x);
    System.out.println("\n");        
}

Answer 4

声明count1＆amp;循环中的chars_not_x变量，或者在每次迭代时将count1和chars_not_x重新初始化为0。

分别分析字符串的每个单词

4 个答案: