Question

我需要做作业才能获得阵列中最“流行”的数字（最高频率的数字），如果有多个数字具有相同数量的节目，请随机获取一些数字。经过三个多小时的尝试，并在网上搜索，这就是我得到的：

public int getPopularNumber(){
        int count = 1, tempCount;
        int popular = array[0];
        int temp = 0;
        for ( int i = 0; i < (array.length - 1); i++ ){
          if ( _buses[i] != null )
            temp = array[i]; 
            tempCount = 0;
            for ( int j = 1; j < _buses.length; j++ ){
              if ( array[j] != null && temp == array[j] ) 
                tempCount++;
            }           
            if ( tempCount > count ){
              popular = temp;
              count = tempCount;
            }
          }
          return popular;
}

此代码有效，但未考虑重要情况 - 如果有多个具有相同节目数的数字。然后它就是第一个。例如：int[]a = {1, 2, 3, 4, 4, ,5 ,4 ,5 ,5};代码将首先显示为4，并且它不应该是随机的。另一件事 - 因为它是家庭作业我不能使用ArrayList / maps和我们仍然没有学习的东西。任何帮助将不胜感激。

Answer 1

由于他们没有给你任何时间复杂性边界，你可以通过扫描阵列N ^ 2次来“暴力破解”问题。（免责声明，这是最直观的方式，在内存和CPU方面不是最快或最有效的。）

这是一些伪代码：

创建另一个与原始数组大小相同的数组，这将是“occurrence数组”
将其元素归零
对于原始数组中的每个索引i，迭代原始数组，并在每次扫描找到存储在{{1}中的值的重复项时，在i处将事件数组中的元素递增在原始数组中。
查找发生数组中的最大值
返回原始数组中该索引中存储的值

通过这种方式，您可以模仿仅使用另一个数组的地图。

Answer 2

如果您不被允许使用收藏，那么您可以尝试以下代码：

public int getPopularNumber(){
    int inputArr[] = {1, 2, 3, 4, 4, 5 ,4 ,5 ,5}; // given input array
    int[] tempArr = new int[inputArr.length];
    int[] maxValArr = new int[inputArr.length];

    // tempArr will have number as index and count as no of occurrence
    for( int i = 0 ; i < inputArr.length ; i++){
        tempArr[inputArr[i]]++;     
    }

    int maValue = 0;
    // find out max count of occurrence (in this case 3 for value 4 and 5) 
    for( int j = 0 ; j < tempArr.length ; j++){
        maValue = Math.max(maValue, tempArr[j]);        
    }

    int l =0;
    // maxValArr contains all value having maximum occurrence (in this case 4 and 5)
    for( int k = 0 ; k < tempArr.length ; k++){
        if(tempArr[k] == maValue){
            maxValArr[l] = k;
            l++;
        }       
    }

    return maxValArr[(int)(Math.random() * getArraySize(maxValArr))];

}

private int getArraySize(int[] arr) {
    int size = 0;
    for( int i =0; i < arr.length ; i++){
        if(arr[i] == 0){
            break;
        }
        size++;
    }
    return size;
}

Answer 3

地狱很难：D 经过一番尝试，我想我已经拥有它（如果有2个数字具有相同的频率，它将首先返回）：

    int mostPopNumber =0;
    int tmpLastCount =0;

    for (int i = 0; i < array.length-1; i++) {
        int tmpActual = array[i];
        int tmpCount=0;
        for (int j = 0; j < array.length; j++) {
            if(tmpActual == array[j]){
                 tmpCount++;
            }
        }
        // >= for the last one 
        if(tmpCount > tmpLastCount){
            tmpLastCount = tmpCount;
            mostPopNumber = tmpActual;
        }
    }

    return mostPopNumber;

-

哈，你的代码给了我一个想法 - 你不能记住最后一个最受欢迎的号码，顺便说一句，我发现它在那里解决了Find the most popular element in int[] array ：）

编辑 - 经过多年，多年：D，效果很好:) 我使用了2D int 和 Integer 数组 - 你也可以只使用int数组，但你必须制作更多的长度数组并复制实际值，Integer有默认值null，所以更快享受

public static void main(String[] args) {
    //income array
    int[] array= {1,1,1,1,50,10,20,20,2,2,2,2,20,20};

    //associated unique numbers with frequency
    int[][] uniQFreqArr = getUniqValues(array);

    //print uniq numbers with it's frequency
    for (int i = 0; i < uniQFreqArr.length; i++) {
        System.out.println("Number: " + uniQFreqArr[i][0] + "  found : " + uniQFreqArr[i][1]);
    }

    //get just most frequency founded numbers
    int[][] maxFreqArray = getMaxFreqArray(uniQFreqArr);

    //print just most frequency founded numbers
    System.out.println("Most freq. values");
    for (int i = 0; i < maxFreqArray.length; i++) {
        System.out.println("Number: " + maxFreqArray[i][0] + "  found : " + maxFreqArray[i][1]);
    }

    //get some of found values and print
    int[] result = getRandomResult(maxFreqArray);
    System.out.println("Found most frequency number: " + result[0] + " with count: " + result[1]);
}

//get associated array with unique numbers and it's frequency
static int[][] getUniqValues(int[] inArray){
    //first time sort array
    Arrays.sort(inArray);

    //default value is null, not zero as in int (used bellow)
    Integer[][] uniqArr = new Integer[inArray.length][2];

    //counter and temp variable
    int currUniqNumbers=1;
    int actualNum = inArray[currUniqNumbers-1];

    uniqArr[currUniqNumbers-1][0]=currUniqNumbers;
    uniqArr[currUniqNumbers-1][1]=1;

    for (int i = 1; i < inArray.length; i++) {
        if(actualNum != inArray[i]){
            uniqArr[currUniqNumbers][0]=inArray[i];
            uniqArr[currUniqNumbers][1]=1;
            actualNum = inArray[i];
            currUniqNumbers++;
        }else{
            uniqArr[currUniqNumbers-1][1]++;
        }
    }

    //get correctly lengthed array
    int[][] ret = new int[currUniqNumbers][2];
    for (int i = 0; i < uniqArr.length; i++) {
        if(uniqArr[i][0] != null){
            ret[i][0] = uniqArr[i][0];
            ret[i][1] = uniqArr[i][1];
        }else{
            break;
        }
    }
    return ret;
}

//found and return most frequency numbers
static int[][] getMaxFreqArray(int[][] inArray){
    int maxFreq =0;
    int foundedMaxValues = 0;

    //filter- used sorted array, so you can decision about actual and next value from array
    for (int i = 0; i < inArray.length; i++) {
        if(inArray[i][1] > maxFreq){
            maxFreq = inArray[i][1];
            foundedMaxValues=1;
        }else if(inArray[i][1] == maxFreq){
            foundedMaxValues++;
        }
    }

    //and again copy to correctly lengthed array
    int[][] mostFreqArr = new int[foundedMaxValues][2];
    int inArr= 0;
    for (int i = 0; i < inArray.length; i++) {
        if(inArray[i][1] == maxFreq){
            mostFreqArr[inArr][0] = inArray[i][0];
            mostFreqArr[inArr][1] = inArray[i][1];
            inArr++;
        }
    }
    return mostFreqArr;
}

//generate number from interval and get result value and it's frequency
static int[] getRandomResult(int[][] inArray){
    int[]ret=new int[2];
    int random = new Random().nextInt(inArray.length);

    ret[0] = inArray[random][0];
    ret[1] = inArray[random][1];
    return ret;
}

Answer 4

您是否可以使用字符串操作而不是地图/列表？：d

int[] arr = new int[] { 10, 5, 2, 4, 9, 10, 4, 4, 8, 7, 10};
String highest = ""; // Collecting all highest numbers in a String
int highestCount = 0;
for (int a : arr) {
    int count = 0;
    for (int b : arr) {
        if (b == a)
            count++;    // Getting the number of the current value
    }
    if (count > highestCount) {
        highest = ":" + a + ":";    // new highest count found, overwriting before counted and collected values
        highestCount = count;
    } else
    if (count == highestCount && !highest.contains(":" + a + ":")) { // if value is not in string yet.
        highest += ":" + a + ":";   // values are wrapped in :x: to prevent ":xy:".contains("x") false positive
    }
}
highest = highest.replaceAll("::", ":").replaceAll("^:", "")
            .replaceAll(":$", ""); // remove first and last ":" and replace :: by :
String[] highestNumbers = highest.split(":"); // split string by :
highest = highest.replaceAll(":", ", ");    // optional for output: replace : by ,
int randomOutput = Integer.parseInt(highestNumbers[(int) (Math.random()
            * (highestNumbers.length - 1))]); // Get random number string and parse to Integer
System.out.println("Highest count: " + highestCount + " for Number(s): "
            + highest + " Random number: " + randomOutput);  // output

从数组

4 个答案: