如何有效地检索号的第一位数 [英] How to retrieve the first digit of number efficiently

问题描述

一个显而易见的解决方案是：

One obvious solution is:

int n = 2134;
while(n > 9)
    n /= 10;

这需要线性时间。我们能不能​​做任何更快？

which takes linear time. Could we do any faster?

这是任何速度比线性时间：

Is this any faster than linear time:

char s[100];
sprintf(s, "%d", n);
n = s[0]-'0';

这是其他方法（效率是首要关注的问题）？
我已经看到了<一个href="http://stackoverflow.com/questions/4525121/first-n-digits-of-a-long-number-in-constant-time">this,但我需要找到只有第一个数字。 （另外，我不明白的答案）。

Which are the other ways (efficiency is primary concern)?
I've seen this, except that I need to find only the first digit. (Also, I don't understand the answer).

推荐答案

一些处理器具有计算有多大一个数字是，很快说明（见<一href="http://en.wikipedia.org/wiki/Leading_zero_count">http://en.wikipedia.org/wiki/Leading_zero_count).这可以用于快速选择10的幂，并通过将其除以代替由10重复进行分流，

Some processors have instructions that calculate "how big" a number is, very quickly (see http://en.wikipedia.org/wiki/Leading_zero_count). This can be used to quickly choose a power of 10, and divide by it, instead of dividing by 10 repeatedly.

假设给你一个函数 CLZ ，计算在一个数字的二进制重新presentation（0 ... 32）前导零的位数。然后，你可以使用一个查找表，给出了10的前导零的每个号码适当的电源。

Suppose you are given a function clz that calculates the number of leading zero bits in a number's binary representation (0...32). Then, you can use a lookup table that gives the proper power of 10 for each number of leading zeros.

uint32_t powers_of_10[33] = {
    1000000000, 1000000000,
    100000000, 100000000, 100000000,
    10000000, 10000000, 10000000,
    1000000, 1000000, 1000000, 1000000,
    100000, 100000, 100000,
    10000, 10000, 10000,
    1000, 1000, 1000, 1000,
    100, 100, 100,
    10, 10, 10,
    1, 1, 1, 1, 1
};

int CalcFirstDecimalDigit(uint32_t x)
{
    int leading_zeros = clz(x);
    x /= powers_of_10[leading_zeros];
    if (x >= 10)
        return 1;
    else
        return x;
}

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