C＃-一个特定的int值在数组中出现多少次？ 没有Linq技术

Question

在SO上已经看到许多答案，但是找不到合适的答案。

我需要一种有效的算法或C＃中的方法来计算特定int值出现在array中的次数 。 没有Linq 。 数组的大小> = 100，每个元素不大于100

具有以下代码：

    for (int i = 0; i < 100; i++)   // get each number for counting
    {
       counter = 0;                 // zero counter for next number comparison

       for (int a = 0; a < array.Length; i++) 
       {
            if (i == array[a])
            {
                counter++;
                if (max < counter) max = counter;   // save max-appeared num
            }   
       }
    }

它在测试质询中显示为结果消息“由于超时而终止”。 我想这段代码需要很多时间才能解决。 还有其他选择吗？

Answer 1

你可以利用

每个元素不大于100

并将所有频率声明为一个数组 （只能包含101项目： [0..100] ）：

int[] freqs = new int[101];

foreach (var item in array)
  freqs[item] += 1;

输出：

for (int i = 0; i < freqs.Length; ++i)
  Console.WriteLine("Number {0} appears {1} times", i, freqs[i]);

通常，对于任意大项目，您必须处理dictionary ：

Dictionary<int, int> freqs = new Dictionary<int, int>();

foreach (var item in array) {
  int v;

  if (freqs.TryGetValue(item, out v)) 
    freqs[item] = v + 1;
  else 
    freqs.Add(1);
}

输出（未排序）：

foreach (var pair in freqs)
  Console.WriteLine("Number {0} appears {1} times", pair.Key, pair.Value);

Answer 2

程序代码：

int[] num = { 1, 1, 1, 3, 3, 4, 5, 6, 7, 0 };
int[] count = new int[10];

//Loop through 0-9 and count the occurances
for (int x = 0; x < 10; x++)
    for (int y = 0; y < num.Length; y++)
        if (num[y] == x)
            count[x]++;

//For displaying output only            
for (int x = 0; x < 10; x++)
    Console.WriteLine("Number " + x + " appears " + count[x] + " times");

程序输出：

Number 0 appears 1 times
Number 1 appears 3 times
Number 2 appears 0 times
Number 3 appears 2 times
Number 4 appears 1 times
Number 5 appears 1 times
Number 6 appears 1 times
Number 7 appears 1 times
Number 8 appears 0 times

我知道当您所有的同学都完成他们的学业后，您仍在苦苦挣扎的感觉。 我的代码应该足够简单以供您学习。

Answer 3

您可以避免执行两个循环

// Prepare the test array (this is done before entering the counting loop)
Random rnd = new Random();
int[] array = new int[100]; // 100 or whatever....
for (int i = 0; i < array.Length; i++)
    array[i] = rnd.Next(1, 101);  // 1-100 range

// The max occurences of a number
int maxOccurence = 0;
// The number with the max occurences
int currentMax = 0;

// One bigger to accomodate the 100....
int[] occurrences = new int[array.Length + 1];

// Loop and check on the test array
for (int a = 0; a < array.Length; a++)
{
    // The current number to examine
    int number = array[a];

    // Increment the occurences counter for the number
    occurrences[number]++;

    // Check if we have a new max....
    if (occurrences[number] > maxOccurence)
    {
        // Save the new leader, both the new max occurence and number 
        maxOccurence = occurrences[number];
        currentMax = number;
    }
}

Console.WriteLine(maxOccurence);
Console.WriteLine(currentMax);

但是，这仍然无法检测两个或多个数字是否具有相同的出现次数。

Answer 4

您可以执行以下操作：

var counters = new int[101]; // 0 to 100
var max = 0;
var maxCount = 0;

foreach (var n in input)
{
    counters[n] = counters[n] + 1;
    if (counters[n] > maxCount)
    {
        maxCount = counters[n];
        max = n;
    }
}

return max;