【LeetCode】50. Pow(x, n)

1.题目描述

Implement pow(x, n), which calculates x raised to the power n (xⁿ).

实现 pow(x, n) ，即计算 x 的 n 次幂函数。

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 [−2³¹, 2³¹ − 1]

2.Solutions

public static double pow(double x, int n) {
    if (n == 0)
        return 1;
    if (n < 0) {
        //If n is Integer.MIN_VALUE, the code will have overflow runtime error.
        if (n == Integer.MIN_VALUE) {
            n++;
            n = -n;
            x = 1 / x;
            return x * x * pow(x * x, n / 2);
        }
        n = -n;
        x = 1 / x;
    }
    //偶数判断
    if((n & 1) == 0){
        return pow(x * x, n / 2);
    }else{
        return x * pow(x * x, n / 2);
    }
}

（完）