Question

我要打印数组中重复次数最多的值。如果两个值的重复次数达到最大次数，则打印最大的一个值。我不知道如何打印最大的值。我尝试过此方法，它只是将打印出的重复值最大的值打印在数组中。

int[] a= { 3,2,3,2,2};
        int count = 1, tempCount;
  int repeated = a[0];
  int temp = 0;
  for (int i = 0; i < (a.length - 1); i++)
  {
    temp = a[i];
    tempCount = 0;
    for (int j = 1; j < a.length; j++)
    {
      if (temp == a[j])
        tempCount++;
    }
    if (tempCount > count)
    {
     repeated = temp;
      count = tempCount;
    }
  }
  System.out.println(repeated);

如果假设数组元素为“ 3,2,3,3,2,4,5,4,6,4”，则必须打印4.（3号3次，4号3次）。 ....但是4是最大编号，因此输出为4）。现在如何更改我的代码？

Answer 1

这里：

@material-ui/core/styles

您发现一个“新的”重复值，然后您无条件地分配了该值。

您需要区分两种情况：

repeated = temp;

解决了您的问题！

Answer 2

将此代码中的j更改为等于0

for (int j = 1; j < a.length; j++)

，因为它跳过了数组的第一个元素，这导致3仅被计数两次。如果较大的数字具有相等的计数，则此逻辑也应有所帮助

    int[] a= {3,2,3,2,2, 2, 4, 4, 5 ,5};
        int count = 1, tempCount;
        int repeated = a[0];
       int temp = 0;
       for (int i = 0; i < (a.length - 1); i++)
        {
       temp = a[i];
       tempCount = 0;
      for (int j = 0; j < a.length; j++)
    {
      if (temp == a[j])
        tempCount++;
    }
    if (tempCount ==count )
    {
        if(temp>repeated ){

     repeated = temp;
      count = tempCount;
        }
    }
    if (tempCount > count)
    {
     repeated = temp;
      count = tempCount;
    }
  }
    System.out.println(repeated);
    }
}

编辑，我知道它很懒，但是我将其保留为海报代码格式。

Answer 3

我可能不会只遍历数组多次，而只遍历一次并计算发生的次数。

在这里Map<Integer, Integer>会有所帮助，特别是如果数字可能为负或有“空洞”（即您有[1,2,5,9,1,9,9]之类的东西）。在这里，键将是数字，值将是计数。示例：

Map<Integer,Integer> counts = new HashMap<>();
for(int n : a ) {
  counts.merge(n, 1, (value,increment) -> value + increment );
}

下一步，您可以按计数对条目进行排序，然后取最高记录，或者再次进行迭代，并跟踪条目（如果它们的计数高于当前最大值，并且其键高于当前键）。

Answer 4

这种方法在您的问题下方的评论中被称为 overkill ，因为其他人（但我）建议使用地图。我认为这是一种有效的方法，因为使用了数据结构Map。查看代码中的注释：

public static void main(String[] args) {
    int[] a = { 3, 2, 3, 2, 2 };
    int[] b = { 3, 2, 3, 3, 2, 4, 5, 4, 6, 4 };
    // create a data structure that holds the element and its count
    Map<Integer, Integer> occurrences = new HashMap<Integer, Integer>();
    /*
     * go through the array and store each element found as the key and how often it
     * was found as the value
     */
    for (int n : b) {
        if (occurrences.containsKey(n)) {
            /*
             * if the key is already contained, increment the value (additional occurrence
             * found)
             */
            occurrences.put(n, occurrences.get(n) + 1);
        } else {
            // otherwiese just add the key with value one (first occurrence)
            occurrences.put(n, 1);
        }
    }

    // now you have all the elements with its occurrences, go find the one to be displayed

    // print them once (not necessary for the result)
    occurrences.forEach((key, value) -> System.out.println(key + " : " + value));

    // get the largest number with the largest occurrence count
    int maxValue = 0;
    int maxKey = 0;

    for (int i : occurrences.keySet()) {
        if (occurrences.get(i) > maxValue) {
            // if you find a larger value, set the current key as max key
            maxKey = i;
            maxValue = occurrences.get(i);
        } else if (occurrences.get(i) == maxValue) {
            /*
             * if you find a value equal to the current largest one, compare the keys and
             * set/leave the larger one as max key
             */
            if (i > maxKey) {
                maxKey = i;
            }
        } else {
            // no need for handling a smaller key found
            continue;
        }
    }

    System.out.println("Largest key with largest value is " + maxKey);
}

Answer 5

这是您的问题的解决方案。阅读代码中的注释以获得更多理解。您的代码的复杂度为O（n ^ 2）。下面的代码具有O（n）复杂度，并且速度更快。

 public static void main(String[] args){
    int[] arr = new int[]{3,2,3,3,2,4,5,4,6,4};
    // Assuming A[i] <= 100. Using freq[] to capture the frequency of number
    // freq[] will behave as a Map where index will be the number and freq[index] will be the frequency
    // of that number
    int[] freq = new int[101];

    for(int i=0;i<arr.length;i++){
        int num = arr[i];
        freq[num]++;
    }
    int max = 0;
    int ans = 0;
    for(int i=0;i<freq.length;i++){
        if(freq[i] >= max){
            ans = i;
            max = freq[i];
        }
    }
    System.out.println(ans);
}

打印最大重复值

5 个答案: