问题

以下是第2和第3步的Java代码：

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

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

        String str1 = "abbc";

        // 1. Build a character frequency map from the list of required characters.
        Map<Character, Integer> requiredChars = new HashMap<>();
        for (Character c : "abbc".toCharArray()) {
            requiredChars.put(c, requiredChars.getOrDefault(c, 0) + 1);
        }

        List<String> inputStrings = new ArrayList<>();
        inputStrings.add("abc");
        inputStrings.add("abcdb");
        inputStrings.add("gggh");
        inputStrings.add("klsrtabcabc");
        inputStrings.add("118*90");
        inputStrings.add("ggggggggggggggggggggggggggggggggg");

        List<String> result = new ArrayList<>();

        // 2. Iterate the list of strings:
        for (String str : inputStrings) {
            // 2.1 Build a character frequency map from the string.
            Map<Character, Integer> strChars = new HashMap<>();
            for (Character c : str.toCharArray()) {
                strChars.put(c, strChars.getOrDefault(c, 0) + 1);
            }

            // 2.2 Check if the string has at least the required number of characters.
            boolean isValid = true;
            for (Map.Entry<Character, Integer> entry : requiredChars.entrySet()) {
                char requiredChar = entry.getKey();
                int requiredCount = entry.getValue();
                int actualCount = strChars.getOrDefault(requiredChar, 0);
                if (actualCount < requiredCount) {
                    isValid = false;
                    break;
                }
            }

            // 2.3 If selected, add the string to the result.
            if (isValid) {
                result.add(str);
            }
        }

        // 3. Return the list of results.
        for (String word : result) {
            System.out.println(word);
        }
    }
}

这段代码执行以下操作：

1. 建立所需字符的字符频率映射。
2. 遍历字符串列表，对每个字符串构建字符频率映射，并检查是否包含足够数量的所需字符。
3. 如果字符串符合条件，将其添加到结果列表中。
4. 最后，输出结果列表中的字符串。

package JavaInterview;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.Set;

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

    String str1="abbc";

    //1. Build a character frequency map from the list of required characters.
            
    Map<Character,Integer> hm=new LinkedHashMap<Character,Integer>();
       
    for(Character c : str1.toCharArray()) {
      if(hm.containsKey(c)) {
        hm.put(c, hm.get(c)+1);
      }
                
      else {
        hm.put(c, 1);
      }
    }
  }
}

答案1

得分: 1

Ideally you should try the steps 2 and 3 and come to SO when you are stuck with an issue and you can't find an answer. For now, I will give you my approach - Just a pseudocode:

for(int i=0;i<arr.length;i++)
{
     Map<Character,Integer> temp = hm;
     for(Character c : arr[i].toCharArray())
     {
         if(!temp.containsKey(c) or temp[c]<=0)
            break the loop;
         else
            temp[c]--;
     }
     if(end of arr[i] charArray)
            answer.add(arr[i]);
}

for(int i=0;i<arr.length;i++)
{
Map<Character,Integer> temp = hm;
for(Character c : arr[i].toCharArray())
{
if(!temp.containsKey(c) or temp[c]<=0)
break the loop;
else
temp[c]--;
}
if(end of arr[i] charArray)
answer.add(arr[i]);
}