java - 使用 HashMap 查找字符串中第二常见的字符

Question

我试图找到字符串中第二个最常见的字符。下面是我的代码

public class secondCommonAlpha {

    public Character secondCommon(String str)
    {
        Character ans = null;
        int first = 0, second = 0;
        HashMap<Character,Integer> counter = new HashMap<>();
        for(char c: str.toCharArray())
        {
            if(!counter.containsKey(c))
            {
                counter.put(c,1);
            }
            else
            {
                counter.put(c,counter.get(c) + 1);
            }
        }
        System.out.println(counter);
       for (char c: counter.keySet())
       {
           if(counter.get(c) > first)
           {
               second = first;

               first = counter.get(c);
           }
           else
               if(counter.get(c) > second && counter.get(c) < first)
           {
               second = counter.get(c);
               ans = c;
           }
       }
        return ans;
    }

    public static void main(String[] args) {
        secondCommonAlpha o = new secondCommonAlpha();
        System.out.println(o.secondCommon("bananassss"));
    }
}

它在第三次迭代中不起作用。如何修复和优化此代码？

编辑

我使用了两个字符变量来完成工作。

public Character secondCommon(String str)
    {
        Character ans = null;
        int first = 0, second = 0;
        Character firstChar = null,secondChar = null;
        HashMap<Character,Integer> counter = new HashMap<>();
        for(char c: str.toCharArray())
        {
            if(!counter.containsKey(c))
            {
                counter.put(c,1);
            }
            else
            {
                counter.put(c,counter.get(c) + 1);
            }
        }
        System.out.println(counter);
       for (char c: counter.keySet())
       {
           if(counter.get(c) > first)
           {

               second = first;
               secondChar = firstChar;
               firstChar = c;

               first = counter.get(c);
           }
           else
               if(counter.get(c) > second && counter.get(c) < first)
           {
               second = counter.get(c);

               secondChar = c;
           }
       }
        return secondChar;
    }

现在工作正常，但我想这段代码可以优化。

Answer 1

在这里，我编写了一个通用实现，它将为您提供 n 个最常见的字符。在您的情况下，您可以指定 2，如示例 main 所示。

答案是通过按出现次数(条目的值)对映射条目集进行降序排序，然后从列表开头选取 n-1(从零开始)索引来完成的。

public static char nMostCommon(String str, int n) {
    Map<Character, Long> counter = str.chars()
        .mapToObj(c -> (char) c)
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
    if (counter.size() < n) {
        throw new IllegalArgumentException("Not enough different characters.");
    }
    return counter.entrySet().stream()
        .sorted(Comparator.comparing(Map.Entry::getValue, Comparator.reverseOrder()))
        .map(Map.Entry::getKey)
        .collect(Collectors.toList())
        .get(n - 1);
}

public static void main(String[] args) {
    System.out.println(nMostCommon("bananassss", 2));
}