Question

我正在编写一个代码，以按固定大小 4个数组的形式打印出最频繁出现的数字，该事件至少出现3次。

例如：

int arr[4] = {9,2,9,9}
Number that occurs at least 3 times is: 9

int arr[4] = {9,9,9,9}
Number that occurs at least 3 times is: 9

int arr[4] = {-1,3,3,3}
Number that occurs at least 3 times is: 3

int arr[4] = {9,-1,9,9}
Number that occurs at least 3 times is: 9

int arr[4] = {1,1,2,2}
Number that occurs at least 3 times is: None

我尝试过：

int arr[4] = {9,2,9,9};
int i;
int count;
int value;
int first;

int length = sizeof(arr) / sizeof(int);
int count = 0;
for (i=1;i<length-1;i++) {
    first = arr[0];
    if (first == arr[i]) {
        value = arr[i];
        count +=1;
    }
}
if (count >= 3) {
    printf("Number that occurs at least 3 times is: %d",value);
}
else {
    printf("Number that occurs at least 3 times is: None");
}

但是我得到了输出：

Number that occurs at least 3 times is: None

应该是：

Number that occurs at least 3 times is: 9

我错过了什么？希望对此有所帮助。

Answer 1

因为您的for循环仅运行2次，计数最多为2。

Answer 2

首先，您必须修复循环以比较最后一个元素：

for (i=1;i<length;i++) {
    ...
}

但是，如果找不到元素，则第一个元素可能不包含最频繁的数字，而其他三个元素却包含。因此，您必须检查其他三个数字是否相同：

if (count >= 3) {
    printf("Number that occurs at least 3 times is: %d",value);
} else if (arr[1] == arr[2] && arr[1] == arr[3])
    printf("Number that occurs at least 3 times is: %d",arr[1]);
} else {
    printf("Number that occurs at least 3 times is: None");
}

这应该涵盖所有情况。

在数组中查找出现至少3次的最频繁元素

2 个答案: