从 Java 中的数字中删除数字

Remove digits from a number in Java

提问by user3159331

How do I remove the first digit of an integer?

如何删除整数的第一个数字？

My input is an integer (for example i = 123456789).

我的输入是一个整数（例如 i = 123456789）。

I then want to remove the first digit, so that i equals 23456789.

然后我想删除第一个数字，以便 i 等于 23456789。

回答by Christian

Here is one way to do it:

这是一种方法：

• Convert it to String
• Take the substring without the first "digit"
• Convert it to int

• 将其转换为 String
• 取没有第一个“数字”的子串
• 将其转换为 int

---

---

Code:

代码：

public static void main(String[] args)
{
    int x = 123456789;

    String x_str = Integer.toString(x);

    int new_x = Integer.parseInt(x_str.substring(1));

    System.out.println(new_x);
}

Output:

输出：

23456789

---

---

Note:This can be done in one line with

注意：这可以在一行中完成

int x = 123456789;
int new_x = Integer.parseInt(Integer.toString(x).substring(1));

Edit:

编辑：

To handle negative-case, check if number is positive or integer:

要处理负情况，请检查 number 是正数还是整数：

int new_x = Integer.parseInt(x > 0 ? 
    Integer.toString(x).substring(1) : Integer.toString(x).substring(2));

回答by Gene

If you want to avoid the string conversion, you can find the high digit and subtract it.

如果你想避免字符串转换，你可以找到高位并减去它。

public static void main(String[] args) {
    int x = 123456789;
    System.out.println("x = " + x);
    int hi = x, n = 0;
    while (hi > 9) {
        hi /= 10;
        ++n;
    }
    for (int i = 0; i < n; i++) hi *= 10;
    x -= hi;
    System.out.println("x with high digit removed = " + x);
}

回答by Evgeniy Dorofeev

try this

尝试这个

n = n % (int) Math.pow(10, (int) Math.log10(n));

回答by Bohemian

Here's the one-line, purely numeric solution:

这是单行的纯数字解决方案：

i %= (int) Math.pow(10, (int) Math.log10(i));

回答by Darren Gilroy

Alternate approach:

替代方法：

int stripLeading(int i) {
  if(i > 0) {
    return i - (int)Math.pow(10, (int)Math.log10(i));
  } else if(i > 0) {
    return i + (int)Math.pow(10, (int)Math.log(-i+1));
  } else {
    return 0;
  }
}

回答by BRPocock

I think I remember the string-free version of this … although I totally agree with @Christian as how I would do it…

我想我记得这个的无字符串版本......虽然我完全同意@Christian我会怎么做......

NOTE: as @Darren Gilroy pointed out, one must consider negatives and zero spocially, and my function fails to do so.

注意：正如@Darren Gilroy 指出的那样，人们必须特别地考虑负数和零，而我的功能却没有这样做。

Of course %is a better solution also.

当然%也是更好的解决方案。

public static void main (String [] argv)
{
     final int x = 123456789;
     int newX = x;

     /* How many digits are there? */
     final double originalLog = Math.floor (Math.log10 (x));

     /* Let's subtract 10 to that power until the number is smaller */
     final int getRidOf = (int)Math.pow (10, originalLog);
     while (originalLog == Math.floor (Math.log10 (newX)))
     { newX -= getRidOf; }

     System.out.println (newX);
}

Poor profiling attempt:

糟糕的分析尝试：

Looping the above function without the printlnfor 20,000,000,000 repeats in a forloop:

在没有printlnfor 20,000,000,000 重复的情况下循环上述函数for：

real    0m9.943s
user    0m9.890s
sys     0m0.028s

The same with Christian's far-easier-to-understand and perfectly functionable version, but for only 200,000,000 repeats (because I'm lazy and got tired of waiting):

与 Christian 的更容易理解且功能完善的版本相同，但仅重复 200,000,000 次（因为我很懒并且厌倦了等待）：

real    0m18.581s
user    0m17.972s
sys     0m0.574s

So one might argue that constructing the String objects is probably slowing it down by roughly 200×, but that isn't a really finely-tuned profiling set-up.

因此，有人可能会争辩说，构造 String 对象可能会使其速度降低大约 200 倍，但这并不是一种真正经过微调的分析设置。

回答by Hungry Blue Dev

If you want to go for simpler methods and without using String, then here's my simple take:

如果您想采用更简单的方法而不使用String，那么这是我的简单做法：

1. Count number of digits int the integer.
2. Divide the intby 10^n. nis the number of digits.
3. Obtain absolute value of the result. //In case of (-)ve numbers.

1. 计算整数的位数。
2. 划分int的10^n。n是位数。
3. 获得结果的绝对值。//如果是 (-)ve 个数字。

For example

例如

int i = 123456789;
int n = getDigitCount(i);
int r = Math.abs(i / (int)Math.pow(10,n)); //r stores result.

And you'd require this method:

你需要这个方法：

int getDigitCount(int num)
{
    int c = 0;
    while(num > 0){
        num/=10;
        c++;
    }
    return c;
}