0050. Pow(x, n) (M)

Pow(x, n) (M)

题目

mplement pow(x, n), which calculates x raised to the power n ((x^n)).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, within the range ([−2^{31}, 2^{31} − 1])

---

题意

手动实现乘方运算。

思路

方法参考快速幂(二分幂)。

---

代码实现

Java

递归

class Solution {
    public double myPow(double x, int n) {
        if (Double.compare(x, 1.0) == 0) {
            return 1.0;
        }
        
        double pow = binaryPow(x, n);
        return n < 0 ? 1 / pow : pow;
    }

    private double binaryPow(double x, int n) {
        if (n == 0) {
            return 1;
        }

        double temp = binaryPow(x, n / 2);
        return n % 2 == 0 ? temp * temp : temp * temp * x;
    }
}

迭代

class Solution {
    public double myPow(double x, int n) {
        if (Double.compare(x, 1.0) == 0) {
            return 1.0;
        }

        double pow = 1.0;
        // 无论正数负数，除以2相当于将其绝对值的二进制右移一位
        for (int i = n; i != 0; i /= 2) {
            // 无论正数负数，不是偶数说明其绝对值二进制末位为1
            if (i % 2 != 0) {
                pow *= x;
            }
            x *= x;
        }

        return n < 0 ? 1 / pow : pow;
    }
}

JavaScript

递归

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function (x, n) {
  if (n === 0) {
    return 1
  }

  let pow = myPow(x, Math.trunc(n / 2))
  if (n % 2 === 0) {
    return pow * pow
  } else if (n > 0) {
    return pow * pow * x
  } else {
    return (pow * pow) / x
  }
}

迭代

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function (x, n) {
  let pow = 1
  let isMinus = n < 0
  while (n !== 0) {
    if (n % 2 !== 0) {
      pow *= x
    }
    x *= x
    n = Math.trunc(n / 2)
  }
  return isMinus ? 1 / pow : pow
}