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在至少出现3次的数组中查找最频繁的元素

arrays c

Maxxx · 技术社区 · 7 年前

4的数组 出现 至少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");
}

Number that occurs at least 3 times is: None

Number that occurs at least 3 times is: 9

我错过了什么?希望您能在这方面给予帮助。

2 回复 | 直到 7 年前

Ctx 7 年前

首先,您必须修复循环以比较最后一个元素:

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

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");
}