我正在尝试编写一个函数,它将数组的大小和由数字组成的 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示例,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.
最佳答案
嵌套循环方法没有任何意义,您只需查看每个数字一次即可计数。当然,用数组来保存计数器更有意义:
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]);
}
关于c - 在C中的int数组中查找数字的频率,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39245576/