Question

我尝试使用以下代码检查数字是否为回文：

unsigned short digitsof (unsigned int x)
{
    unsigned short n = 0;
    while (x)
    {
        x /= 10;
        n++;
    }
    return n;
}

bool ispalindrome (unsigned int x)
{
    unsigned short digits = digitsof (x);

    for (unsigned short i = 1; i <= digits / 2; i++)
    {
        if (x % (unsigned int)pow (10, i) != x % (unsigned int)pow (10, digits - 1 + i))
        {
            return false;
        }
    }
    return true;
}

但是，以下代码无法检查回文 - 即使数字是回文，也总是返回false。

有人可以指出错误吗？

（请注意：我没有兴趣将它变成一个字符串并反转它以查看问题所在：相反，我很想知道上面代码中的错误位置。）

Answer 1

我个人只是从数字中构建一个字符串，然后将其视为正常的回文检查（检查上半部分中的每个字符是否与length()-index处的字符匹配）。

Answer 2

问题在于：

x % (unsigned int)pow (10, i)

让我们试试：

x =504405
i =3

SO I want 4.

x % 10^3 => 504405 %1000 => 405 NOT 4

怎么样

x / (unsigned int)pow (10, i -1) % 10

Answer 3

x % (unsigned int)pow (10, i)不是第i位。

Answer 4

仅供参考！以下两个功能对我有用：

double digitsof (double x)
{
    double n = 0;
    while (x > 1)
    {       
        x /= 10;
        n++;
    }
    return n;
}

bool ispalindrome (double x)
{
    double digits = digitsof (x);
    double temp = x;
    for(double i = 1; i <= digits/2; i++)
    {
        float y = (int)temp % 10;
        cout<<y<<endl;
        temp = temp/10;

        float z = (int)x / (int)pow(10 , digits - i);
        cout<<(int)z<<endl;
        x = (int)x % (int)pow(10 , digits - i);

        if(y != z)
            return false;

    }
    return true;        
}

Answer 5

代码，以检查给定编号是否是回文，是否在JAVA中

I have updated your code, see below..

    public class WeekFragment extends Fragment {
    //Create a global View instance like below
    View previousView;

        public WeekFragment() {
            // Required empty public constructor
        }

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container,
                                 Bundle savedInstanceState) {
            // Inflate the layout for this fragment
            return inflater.inflate(R.layout.fragment_week, container, false);
        }

        @RequiresApi(api = Build.VERSION_CODES.JELLY_BEAN)
        @Override
        public void onViewCreated(@NonNull View view, @Nullable Bundle savedInstanceState) {
            super.onViewCreated(view, savedInstanceState);

            TableLayout timeTable = getActivity().findViewById(R.id.timeTable);
            int count = timeTable.getChildCount();
            for (int i = 0; i < count; i++) {
                View v = timeTable.getChildAt(i);
                if (v instanceof TableRow) {
                    TableRow row = (TableRow) v;
                    int rowCount = row.getChildCount();
                    for (int r = 0; r < rowCount; r++) {
                        final View v2 = row.getChildAt(r);
                        if (v2 instanceof ImageView) {
                            final ImageView b = (ImageView) v2;

                        b.setOnClickListener(new View.OnClickListener() {
                            @Override
                            public void onClick(View v) {
                            b.setBackground(getResources().getDrawable(R.drawable.cell_highlight));
//Check for the first time for null value
    if(previousView!=null&&previousView.equals(b)){
/*You can make cell normal here by adding b.setBackgroundResource(R.drawable.normal);*/
switchActivities();
}
    previousView=null;
    //Assign new view
    previousView=b;
                            }
                        });
                        }
                    }
                }
            }
        }
    }

Answer 6

我已经用自己的解决方案完成了此操作，这些解决方案受这些条件的限制

请勿将int转换为字符串。
请勿使用任何辅助功能。

var inputNumber = 10801

var firstDigit = 0

var lastDigit = 0

var quotient = inputNumber



while inputNumber > 0 {

    lastDigit = inputNumber % 10

    var tempNum = inputNumber

    var count = 0

    while tempNum > 0 {

        tempNum = tempNum / 10

        count = count + 1

    }

    var n = 1

    for _ in 1 ..< count {

        n = n * 10

    }

    firstDigit = quotient / n

    if firstDigit != lastDigit {

        print("Not a palindrome :( ")

        break

    }

    quotient = quotient % n

    inputNumber = inputNumber / 10

}

if firstDigit == lastDigit {

    print("It's a palindrome :D :D ")

}

检查一个数字是否是回文

6 个答案: