java - 如何在不使用映射的情况下计算字符串在数组中出现的次数？

Question

我正在尝试计算字符串在数组中出现的次数，但我无法使用任何类型的 Map。根据我的研究，我尝试了以下方法，希望它能起作用:

int count = Collections.frequency(strings, search);

但是，当我运行程序时出现错误。该错误是空指针异常错误，我不确定如何修复该错误或发生这种情况的原因。关于我的代码有什么问题有什么想法吗？

编辑:这是我的完整代码。我找不到问题出在哪里。也许其他人可以找到它 导入java.util.*;

公开课 Lab9{

 public static void main(String[] args){
     Scanner kb =  new Scanner (System.in);
     String name = null;
     List<String> strings =null;
     System.out.println("How many strings do you want to enter?:");
     int size = kb.nextInt();

     List<String> String = getStrings(size,name);
     System.out.println(String);
     System.out.println("What string would you like to search for?:");
     String search = kb.next();

     int numofTimes = countValues (search,strings);
     System.out.print("That word appears "+numofTimes+" times in the array.");


 }
 public static List<String> getStrings(int size,String name) {
     Scanner kb = new Scanner (System.in);
     int count = 0;
     List<String> strings = new ArrayList<String>();
     while (count != size) {
         System.out.println("Enter a string:");
         name = kb.nextLine();
         strings.add(name);
         count = count + 1;
          }

return strings;

 }
 public static int countValues (String search, List<String> strings){

     int count = Collections.frequency(strings , search);



     return count;
 }



 }

Answer 1

您可以通过数组进行线性搜索

    String strings[] = {"A","B",null,"C","A",null}; // the array contains nulls
    String search = "A";
    int count = 0;
    for (int i = 0;i< strings.length ;i++ ) 
    {
        if(strings[i] != null)
        {
            if(strings[i].equals(search))
            {
                count++;
            }
        }                   
    }
    System.out.println(count);