迟到的人:

下面是一个尽可能处理y < 0的解。

It uses a result of intmax_t for maximum range. There is no provision for answers that do not fit in intmax_t. powjii(0, 0) --> 1 which is a common result for this case. pow(0,negative), another undefined result, returns INTMAX_MAX intmax_t powjii(int x, int y) { if (y < 0) { switch (x) { case 0: return INTMAX_MAX; case 1: return 1; case -1: return y % 2 ? -1 : 1; } return 0; } intmax_t z = 1; intmax_t base = x; for (;;) { if (y % 2) { z *= base; } y /= 2; if (y == 0) { break; } base *= base; } return z; }

这段代码使用了一个永久循环for(;;)，以避免在其他循环解决方案中常见的最终基数*=基数。这个乘法是1)不需要的，2)可能是int*int溢出，也就是UB。

更一般的解决方案考虑负指数

private static int pow(int base, int exponent) {

    int result = 1;
    if (exponent == 0)
        return result; // base case;

    if (exponent < 0)
        return 1 / pow(base, -exponent);
    int temp = pow(base, exponent / 2);
    if (exponent % 2 == 0)
        return temp * temp;
    else
        return (base * temp * temp);
}

下面是Java中的方法

private int ipow(int base, int exp)
{
    int result = 1;
    while (exp != 0)
    {
        if ((exp & 1) == 1)
            result *= base;
        exp >>= 1;
        base *= base;
    }

    return result;
}

迟到的人:

下面是一个尽可能处理y < 0的解。

It uses a result of intmax_t for maximum range. There is no provision for answers that do not fit in intmax_t. powjii(0, 0) --> 1 which is a common result for this case. pow(0,negative), another undefined result, returns INTMAX_MAX intmax_t powjii(int x, int y) { if (y < 0) { switch (x) { case 0: return INTMAX_MAX; case 1: return 1; case -1: return y % 2 ? -1 : 1; } return 0; } intmax_t z = 1; intmax_t base = x; for (;;) { if (y % 2) { z *= base; } y /= 2; if (y == 0) { break; } base *= base; } return z; }

这段代码使用了一个永久循环for(;;)，以避免在其他循环解决方案中常见的最终基数*=基数。这个乘法是1)不需要的，2)可能是int*int溢出，也就是UB。

除了Elias的答案，当使用有符号整数实现时，会导致未定义行为，当使用无符号整数实现时，会导致高输入的不正确值，

下面是平方求幂的修改版本，它也适用于有符号整数类型，并且不会给出错误的值:

#include <stdint.h>

#define SQRT_INT64_MAX (INT64_C(0xB504F333))

int64_t alx_pow_s64 (int64_t base, uint8_t exp)
{
    int_fast64_t    base_;
    int_fast64_t    result;

    base_   = base;

    if (base_ == 1)
        return  1;
    if (!exp)
        return  1;
    if (!base_)
        return  0;

    result  = 1;
    if (exp & 1)
        result *= base_;
    exp >>= 1;
    while (exp) {
        if (base_ > SQRT_INT64_MAX)
            return  0;
        base_ *= base_;
        if (exp & 1)
            result *= base_;
        exp >>= 1;
    }

    return  result;
}

使用该函数的注意事项:

(1 ** N) == 1
(N ** 0) == 1
(0 ** 0) == 1
(0 ** N) == 0

如果将发生任何溢出或换行，则返回0;

I used int64_t, but any width (signed or unsigned) can be used with little modification. However, if you need to use a non-fixed-width integer type, you will need to change SQRT_INT64_MAX by (int)sqrt(INT_MAX) (in the case of using int) or something similar, which should be optimized, but it is uglier, and not a C constant expression. Also casting the result of sqrt() to an int is not very good because of floating point precission in case of a perfect square, but as I don't know of any implementation where INT_MAX -or the maximum of any type- is a perfect square, you can live with that.

O(log N)的解决方案在Swift…

// Time complexity is O(log N)
func power(_ base: Int, _ exp: Int) -> Int { 

    // 1. If the exponent is 1 then return the number (e.g a^1 == a)
    //Time complexity O(1)
    if exp == 1 { 
        return base
    }

    // 2. Calculate the value of the number raised to half of the exponent. This will be used to calculate the final answer by squaring the result (e.g a^2n == (a^n)^2 == a^n * a^n). The idea is that we can do half the amount of work by obtaining a^n and multiplying the result by itself to get a^2n
    //Time complexity O(log N)
    let tempVal = power(base, exp/2) 

    // 3. If the exponent was odd then decompose the result in such a way that it allows you to divide the exponent in two (e.g. a^(2n+1) == a^1 * a^2n == a^1 * a^n * a^n). If the eponent is even then the result must be the base raised to half the exponent squared (e.g. a^2n == a^n * a^n = (a^n)^2).
    //Time complexity O(1)
    return (exp % 2 == 1 ? base : 1) * tempVal * tempVal 

}