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Amazon VO 真题,我是如何拿到Amazon offer的


Implement the function pow(x, n), which calculates x^n.

Note that you cannot use any existing functions that implement pow, as that would trivialize the problem.

Example 1: Input: x = 2, n = 5 output = 32


When presented with the interview question of implementing a pow(x, n) function, which calculates x to the power of n, my first step would be to clarify the problem's constraints and edge cases. This includes inquiring about the range and type of x, as well as handling scenarios where n is negative or zero.

我接着会提到一种更高效的方法——快速幂算法。快速幂算法的核心思想是分治,将x^n分解为若干个小的部分,利用n的二进制表示来降低计算的复杂度。具体来说,我们可以将n表示为二进制数,并利用x^n = x^(2^k) * x^(2^(k-1)) * ... * x^(2^0)的性质,其中kn的位数。这样,我们只需要对n的每一个二进制位进行检查,如果该位为1,则将当前的x^(2^i)i为该位的位置)乘到结果上。这种方法将时间复杂度降低到了O(log n)。

I would start by outlining my approach to the interviewer, beginning with a brute-force method and then introducing a more efficient solution. The straightforward method involves multiplying x by itself n times, but this approach has a time complexity of O(n), which can be impractical for large values of n.

Next, I would discuss a more sophisticated method known as the fast exponentiation algorithm. The essence of this algorithm is a divide-and-conquer strategy that breaks down the computation of x^n into smaller chunks, taking advantage of the binary representation of n to reduce computational complexity. Specifically, it leverages the fact that x^n = x^(2^k) * x^(2^(k-1)) * ... * x^(2^0), where k is the number of bits in n. This means we only need to consider each bit of n, multiplying the result by x^(2^i) (where i is the position of the bit) if that bit is set to 1. This algorithm reduces the time complexity to O(log n).


Here's the pseudocode for this solution:


  1. 处理n为负数的情况,这时候我们需要计算1/x-n次幂。
  2. 初始化结果result为1,因为任何数的0次幂都是1。
  3. 使用current_product变量跟踪x的幂次方的值,这个值在每次循环时平方,以此实现快速幂。
  4. 通过n的二进制表示来决定是否将current_product乘到result上。
  5. 通过n = n // 2来实现对n的二进制位的遍历。


While explaining this pseudocode, I would emphasize several key points:

  1. Handling negative n by computing the power of 1/x to the -n.
  2. Initializing result to 1 because any number to the power of 0 is 1.
  3. The use of a current_product variable to keep track of the power of x, which is squared in each iteration to achieve fast exponentiation.
  4. Deciding to multiply current_product with result based on the binary representation of n.
  5. Iterating through the bits of n by dividing n by 2 in each step.

This approach not only allows me to efficiently solve the problem but also demonstrates clear thinking and code implementation skills to the interviewer.

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