计算数组中的重复值

Question

我想编写一个程序来计算数组中的重复项.如果两个相同的数字,代码可以工作.但是,如果有三个或更多相同的数字,则会出错.我该怎么办呢？

public class Duplicate
    {
      public static void main(String[] args)
      {
        int[] list = new int[]{1,2,3,4,5,6,7,8,8,8,9,10};

        int sum = 0;
        for(int count=1; count<list.length; count++)
        {
          if(list[count-1]==list[count])
          {
            sum = list[count-1] + list[count];
            System.out.println("Duplicate found: " + list[count] + " " + "Sum of the duplicate value is " +sum);
          }
        }
      }
    }

Answer 1

这是一种Java-8风格的功能方法:

int[] array = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 8, 8, 9, 10 };

// Create a Stream<Integer> from your data
IntStream.of(array)
         .boxed()

// Group values into a Map<Integer, List<Integer>>
         .collect(Collectors.groupingBy(i -> i))

// Filter out those map values that have only 1 element in their group
         .entrySet()
         .stream()
         .filter(e -> e.getValue().size() > 1)

// Print the sum for the remaining groups
         .forEach(e -> {
             System.out.println(
                 "Duplicates found for : " + e.getKey() +
                 " their sum being : " + e.getValue()
                                          .stream()
                                          .collect(Collectors.summingInt(i -> i)));
         });

对于您的输入,这会产生:

Duplicates found for : 8 their sum being : 24

这个解决方案的好处在于它适用于无序的解决方案int[].例如......

int[] array = new int[] { 1, 10, 3, 2, 3, 4, 5, 8, 6, 7, 8, 8, 8, 9, 10 };

输出将是......

Duplicates found for : 3 their sum being : 6
Duplicates found for : 8 their sum being : 32
Duplicates found for : 10 their sum being : 20