java - 如何获取数组中最常见的字符？

Question

假设我有一个这样的整数数组:{5,3,5,4,2}

我有一个返回最常见字符的方法

public int highestnumber(String[] num) {

        int current_number = Integer.parseInt(num[0]);
        int counter = 0;
        for (int i = 1; i < num.length; ++i) {
            if (current_number == Integer.parseInt(num[i])) {
                ++counter;

            } else if (counter == 0) {
                current_number = Integer.parseInt(num[i]);
                ++counter;

            } else {
                --counter;

            }
        }

        return current_number;
    }

但是如果我有多个公共(public)字符，那么我需要获取最接近一个 (1) 的数字，就像我有一个这样的数组一样:{5,5,4,4,2};

那么该方法应该返回 4，我应该为此做什么？

Answer 1

根据我对你问题的理解，

你要做的是，

1. Create ArrayList from your int[]
2. Use HashMap for find duplicates, which one is unique
3. Sort it as Ascending order, 
4. First element is what you want..

编辑:回答您的问题

int[] arr = {5, 4, 5, 4, 2};
ArrayList<Integer> resultArray = new ArrayList<Integer>();
Set<Integer> set = new HashSet<Integer>();

for (int i = 0; i < arr.length; i++)
    {
     if (set.contains(arr[i]))
    {
     System.out.println("Duplicate value found at index: " + i);
     System.out.println("Duplicate value: " + arr[i]);
         resultArray.add(arr[i]);
    }
    else
    {
    set.add(arr[i]);
    }
   }
Collections.sort(resultArray);

for (int i = 0; i < resultArray.size(); i++)
{
Log.e("Duplicate Values:", resultArray.get(i) + "");
}

您的需求是，

int values = resultArray.get(0);