需要在java数组中找到重复整数及其计数

Question

我使用下面的代码，但输出不正确

// 1,3,4,5,6,3,2,4,6,7,9,4,12,3,4,6,8,9,7,6,43,2,4 ，7,7,5,2,1,3,4,6,311,1

    public static void repeat(){

        int count[] = new int[]{1,3,4,5,6,3,2,4,6,7,9,4,12,3,4,6,8,9,7,6,43,2,4,7,7,5,2,1,3,4,6,311,1};
        int [] done ;
        for(int i = 0; i < count.length; i++) {

            int a = count[i] ;
            int counter=0;  
            int j=i+1;

            //System.out.println(a+"--"+""+j);
            for(j=i+1 ; j < count.length; j++){
                if (a == count[j]) {        
                     counter=counter+1;
            }
                System.out.println(a+" is appearing --"+""+counter +" times");
            }

        }   

        }

Answer 1

尝试创建一个键类型为Integer且值相同的Map。然后遍历count数组 - 检查map中是否存在给定值 - 如果没有将其添加为值为1的键，如果它确实存在，则只需将该值增加1.

在此之后浏览地图并获取值为＆gt;的所有键。 1。

 public static void repeat(){

    int count[] = new int[]    {1,3,4,5,6,3,2,4,6,7,9,4,12,3,4,6,8,9,7,6,43,2,4,7,7,5,2,1,3,4,6,311,1};

    Map<Integer,Integer> repetitionCount = new HashMap<>();

    for(Integer i : count) {
       if(repetitionCount.containsKey(i)) {
           Integer prevValue = repetitionCount.get(i);
           repetitionCount.put(i,prevValue+1);
       } else {
           repetitionCount.put(i,1);
       }
    }

    repetitionCount.forEach((key,value) -> {
            if(value>1) {
                System.out.println("Repetition of " + key + " - " + value + " times");
            }
        }
    );

}