Question

我正在尝试编写一个函数，它接受一个数组的大小和一个由数字组成的int数组作为输入，并打印每个数字的频率。

示例输入和输出：

Input: [1,2,2,3,3,3]

Output:
1 occurs 1 times.
2 occurs 2 times
3 occurs 3 times.

这是我的尝试（不是最优雅的）：

void freq(int size, int numArray[]) {
    int one=0, two=0, thr=0, fou=0, fiv=0, six=0, sev=0, eit=0, nin=0;
    int i, j;

    for (i = 0; i < size; i++) {
        for (j = 1; j < size; j++) {
            if (numArray[i] == numArray[j] && numArray[i] == 1) {
                one+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 2) {
                two+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 3) {
                thr+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 4) {
                fou+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 5) {
                fiv+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 6) {
                six+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 7) {
                sev+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 8) {
                eit+=1;
            }
            else if (numArray[i] == numArray[j] && numArray[i] == 9) {
                nin+=1;
            }
        }
    }
    printf("1 occurs %d times.\n", one);
    printf("2 occurs %d times.\n", two);
    printf("3 occurs %d times.\n", thr);
    printf("4 occurs %d times.\n", fou);
    printf("5 occurs %d times.\n", fiv);
    printf("6 occurs %d times.\n", six);
    printf("7 occurs %d times.\n", sev);
    printf("8 occurs %d times.\n", eit);
    printf("9 occurs %d times.\n", nin);
}

这有问题。如果我使用与上面相同的例子，这就是我得到的：

Input: [1,2,2,3,3,3]

Output:
1 occurs 0 times.
2 occurs 4 times.
3 occurs 9 times.

Answer 1

嵌套循环方法毫无意义，您只需要查看每个数字一次就可以计算它。当然，数组更有意义保留计数器：

void freq(int size, const int *numbers)
{
  unsigned int counts[10] = { 0 };
  for(int i = 0; i < size; ++i)
  {
    const int here = numbers[i];
    if(here >= 1 && here <= 9)
      counts[here]++;
  }
  for(int i = 1; i < 10; ++i)
    printf("%d occurs %u times\n", i, counts[i]);
}

Answer 2

你的逻辑有缺陷。你想要的是在数组上迭代一次并检查每个元素。像这样的东西：

for(int i = 0; i < size; i++){
    if (numArray[i] == NUMBER)
       NUMBER_COUNTER ++;
.
.
.

或者，使用原始代码：

void freq(int size, int numArray[]) {
    int one=0, two=0, thr=0;
    int i;

    for (i = 0; i < size; i++) {
            if (numArray[i] == 1) {
                one+=1;
            }
            else if (numArray[i] == 2) {
                two+=1;
            }
            else if (numArray[i] == 3) {
                thr+=1;
            }
    }
    printf("1 occurs %d times.\n", one);
    printf("2 occurs %d times.\n", two);
    printf("3 occurs %d times.\n", thr);

}

我认为更优雅的另一种解决方案是：创建一个包含9个元素的新数组（假设您想要计算9位数的频率......）并增加找到的数字的位置......就像这样：

    void freq(int size, int numArray[]) {
    int freq_arr[size];
    int i;

    for(i = 0; i < size; i++)
        freq_arr[ numArray[i] ] ++;
    for(i = 0; i < size; i++)
        printf("i: %d = %d\n", i, freq_arr[i]);

Answer 3

你需要内循环什么？您需要的是在每次迭代中查看每个数字一次以便对其进行计数。它比你想象的要简单。将您的代码更改为：

void freq(int size, int numArray[]) {
    int one=0, two=0, thr=0, fou=0, fiv=0, six=0, sev=0, eit=0, nin=0;
    int i, j;

    for (i = 0; i < size; i++) {
        if (numArray[i] == 1) 
            one+=1;
        else if (numArray[i] == 2)
            two+=1;
        else if (numArray[i] == 3) 
            thr+=1;      
        else if (numArray[i] == 4) 
            fou+=1;
        else if (numArray[i] == 5) 
            fiv+=1;
        else if (numArray[i] == 6) 
            six+=1;
        else if (numArray[i] == 7) {
            sev+=1;
        else if (numArray[i] == 8) 
            eit+=1; 
        else if (numArray[i] == 9) 
            nin+=1;
}
    printf("1 occurs %d times.\n", one);
    printf("2 occurs %d times.\n", two);
    printf("3 occurs %d times.\n", thr);
    printf("4 occurs %d times.\n", fou);
    printf("5 occurs %d times.\n", fiv);
    printf("6 occurs %d times.\n", six);
    printf("7 occurs %d times.\n", sev);
    printf("8 occurs %d times.\n", eit);
    printf("9 occurs %d times.\n", nin);
}

另一种方法不是将十个int变量保留为计数器，而是保留十个int的数组，如下所示：

void freq(int size, int numArray[]) {

    int i;
    int counters[10];
    //initialize the array's elements to zero
    for (i = 0; i < 10; i++)
        counters[i] = 0;

    for(i = 0; i < size; i++)
    {
        if (numArray[i] = i)
            counters[i]++;
    }
}

Answer 4

void freq(int size, const int numArray[]) {
    int cnt[10] = {0};
    for (int i=0; i<size; i++) {
         if (0 <= numArray[i] && numArray[i] <= 9)
           cnt[numArray[i]]++;
    }
    // cnt = {0, 3, 1, 0, 0, 0, 0, 0, 0, 0}
}

// test
int main(int argc, char* argv[])
{
    int arr[4] = {1,1,1,2};
    freq(4, arr);
    return 0;
}

在C

4 个答案: