试图在 java 中的数组中查找字符串索引

Trying to find index of string in array in java

我试图找到数组中最长字符串的 indexOf,它由字符串 longString 表示。我不断收到 "cannot find symbol" 错误并且不知道如何解决此问题

public class Final21 {
       public static String getLongString(String[] array) {
          int x=0;
          int maxLength = 0;
          String longString = null;
          for (String s : array) {
              if (s.length() > maxLength) {
                  maxLength = s.length();
                  longString = s;
              }
          }
          return longString;
      }

         public static String getShortString(String[] array) {
          int minLength = 0;
          String shortString = null;
          for (String t : array) {
              if (t.length() > minLength) {
                  minLength = t.length();
                  shortString = t;
              }
          }
          return shortString;
      }

      public static void main(String[] args) {
          String[] names = {"bob", "maxwell", "charley", "tomtomjack"};
             String longString = getLongString(names);
                System.out.println("The Longest String is: " + longString + " With The Index Of" + names.indexOf(longString));

             String shortString = getShortString(names);
                System.out.println("The Longest String is: " + shortString + " With The Index Of" );
      }

    }

问题出在names.indexOf(longString)。因为 namesString[] 类型,它是一个数组。此类型 String[] 没有 名为 indexOf 的方法定义。作为替代方案,您可以尝试 java.util.Arrays.asList(theArray).indexOf(o)

因此,要更正您的代码片段,您可以像这样重写它

System.out.println("The Longest String is: " + longString + " With The Index Of" + java.util.Arrays.asList(names).indexOf(longString));

阅读 Javajava.util.Arrays 的文档以温习如何使用 Java API 处理 java 中的数组的知识。

此外,您可以通过修改代码的语义来实现相同的目的。 by Elliott Frisch 已经为您做到了。也读一下..

数组中没有 indexOf 函数,而不是您当前的方法 - 我会 return 来自 getLongString 和 [=16= 的 index ] 方法;首先假设它是第一个元素。如果任何元素更长(或更短),请更新 return 值。喜欢,

public static int getLongString(String[] array) {
    int max = 0;
    for (int i = 1; i < array.length; i++) {
        if (array[i].length() > array[max].length()) {
            max = i;
        }
    }
    return max;
}

public static int getShortString(String[] array) {
    int min = 0;
    for (int i = 1; i < array.length; i++) {
        if (array[i].length() < array[min].length()) {
            min = i;
        }
    }
    return min;
}

那你可以这样称呼它

public static void main(String[] args) {
    String[] names = { "bob", "maxwell", "charley", "tomtomjack" };
    int longString = getLongString(names);
    System.out.println("The Longest String is: " + names[longString] + " With The Index Of " + longString);

    int shortString = getShortString(names);
    System.out.println("The Longest String is: " + names[shortString] + " With The Index Of " + shortString);
}

输出

The Longest String is: tomtomjack With The Index Of 3
The Longest String is: bob With The Index Of 0

indexOf() 的语法是:

public int indexOf(char ch);

它 returns 指定字符在该字符串中第一次出现的索引,如果该字符未出现,则为 -1。

获得所需的输出。

您可以尝试使用以下方法更改您的代码;

public static String getLongString(String[] array) {
     int x=0, maxLength = 0, index = 0;
     String longString = null;

     for (String s : array) {
         if (s.length() > maxLength) 
         {
             maxLength = s.length();
             index++;
             longString = s;
         }
     }
 return "The Longest String is: "+ longString + " With The Index Of :" + index;
}

 public static void main(String[] args) {
       String[] names = {"bob", "maxwell", "charley", "tomtomjack"};
       String longString = getLongString(names);
       System.out.println(longString);

 }