Question

我想编写一个代码，其中我必须将数字与数组的所有元素进行比较。该数字只能等于数组的一个元素或不等于任何元素。我可以使用for循环中的if语句将数字与数组的每个元素进行比较。问题是当我想写＃34;数字不等于数组中的任何元素时。下面显示的代码将执行else语句99或100次，但我需要只执行一次，将数字与所有X [i]进行比较后找不到相同的数字。

for(int i = 0; i < 100; i++)
{
    if(number == X[i])
    {
        cout << "number is equal to one of the numbers in array" << endl;
    }
    else
    {
        cout << "number is not equal to any number in the array" << endl;
    }
}

Answer 1

这段代码应该是正确的：当它找到一个与你要查找的数字相等的数组元素时，会打破将bool变量转换为true的循环。因此，如果它为假（if(!check)），则该数字不在数组中。

bool check = false;

for (int i = 0; i < 100; i++)
{
    if(number == X[i])
    {
        cout<<"number is equal to one of the numbers in array"<<endl;
        check = true;
        break;
    }
}

if (!check)
    cout<<"number is not equal to any number in the array"<<endl;

Answer 2

您可以尝试STL提供的算法。 std::find算法可能就是你要找的。

Answer 3

那是因为您要将“数字”与数组“X”中的每个数字进行比较，并且每次2个数字不时等于你打印那个陈述。

你想要的是更像这样的东西：

bool foundNumber = false;

for(int i=0;i<100;i++){

    if(number==X[i]){

        cout<<"number is equal to one of the numbers in array"<<endl;
        foundNumber = true;
        break; //The "break" command just exits out of the loop and if you already know it's equal, you can just exit the loop

    }
}

//Now that we are out of the loop, we check the "foundNumber" variable
if(foundNumber==false){

    //If "foundNumber" is false, then we can print out that we did not find the number
    cout<<"number is not equal to any number in the array"<<endl;

}

Answer 4

我认为这可能会提供您正在寻找的答案，使用计数器来查明它是否存在多次。注意 - 我没有运行此代码！据我所知，鉴于略微模棱两可的问题，它更具可扩展性，并且应满足您是否需要找到一次显示的值，仅一次的需求，或不止一次。

unsigned int count = 0;
for (int i = 0; i < 100; i++)
{
    if(number == X[i])
    {
         count++;
    }
}

if (count == 1) //Edit with == x, or > or < symbols to change frequency requirements
    cout << "Number exists " << count << " time(s) in array"<<endl;
else
    cout << "Number does not exist only once in array" << endl;

将数字与数字数组进行比较

4 个答案: