-1

我尝试了不同的方法来获取用户的输入序列，但不知道该怎么做。这是我的代码。大小为 N 的数组 A 中的多数元素是出现超过 N/2 次的元素。例如 (3,3,4,2,4,4,2,4,4) 有一个多数元素 (4)，而数组 (3,3,4,2,4,4,2,4) 有没有多数元素。我正在尝试从用户那里获取输入序列。

import java.util.Scanner;

class rough1 {
   public static int arrMajority1( int A[] ) {
      int n = A.length;
      for( int i = 0; i < A.length; i++ ) {
         int c = 1;
         for( int j = i + 1; j < A.length; j++ )
            if( A[ i ] == A[ j ] )
               c = c + 1;
         if( c > ( A.length / 2 )) {
            return A[ i ];
         }
      }
      return -1;
   }

   public static void main(String[] args){
      Scanner input = new Scanner(System.in);
      int A[] = new int [];
      A[] = input.nextInt();
      String employee = "A[]";
      String delims = "[,]";
      String[] tokens = employee.split(delims);
      if (arrMajority1(A) != -1)
         System.out.println("The majority element is " + arrMajority1(A));
      else 
         System.out.println("There is no majority element.");
   }
}
4

1 回答 1

1

这个怎么样？

public static void main(String[] args) throws FileNotFoundException {
    int[] arrayWithMajorityElements = {3,3,4,2,4,4,2,4,4};

    Integer[] majorityElements = getMajorityElements(arrayWithMajorityElements);
    System.out.println(Arrays.toString(majorityElements));
    // result = 4

    int[] arrayWithoutMajorityElements = {3,3,4,2,4,4,2,4};
    majorityElements = getMajorityElements(arrayWithoutMajorityElements);
    System.out.println(Arrays.toString(majorityElements));
    // result = empty array (ie no majory elements)
}

private static Integer[] getMajorityElements(int[] array) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i : array) {
        int count = 0;
        if (map.containsKey(i)) {
            count = map.get(i);
        }
        map.put(i, ++count);
    }

    List<Integer> majorityElements = new ArrayList<>();
    int check = array.length / 2;
    for (Integer i : map.keySet()) {
        if (map.get(i) > check) {
            majorityElements.add(i);
        }
    }
    return majorityElements.toArray(new Integer[majorityElements.size()]);
}

请注意，我假设您可以拥有多个多数元素-如果不能，则必须修改最后一个循环

于 2013-02-20T20:47:24.497 回答