leetcode 50.Pow(x,n) Java

Pow(x,y)

做题博客链接

https://blog.csdn.net/qq_43349112/article/details/108542248

题目链接

https://leetcode-cn.com/problems/powx-n/

描述

实现 pow(x, n) ,即计算 x 的 n 次幂函数(即,xn)。


提示:

-100.0 < x < 100.0
-231 <= n <= 231-1
-104 <= xn <= 104

示例

示例 1:

输入:x = 2.00000, n = 10
输出:1024.00000

示例 2:

输入:x = 2.10000, n = 3
输出:9.26100

示例 3:

输入:x = 2.00000, n = -2
输出:0.25000
解释:2-2 = 1/22 = 1/4 = 0.25

初始代码模板

class Solution {
    
    
    public double myPow(double x, int n) {
    
    
      
    }
}

代码

class Solution {
    
    
    public double myPow(double x, int n) {
    
    
        if (x == 0.0) {
    
    
            return 0.0;
        }

        double res = 1.0;
        long b = n;
        if (n < 0) {
    
    
            x = 1 / x;
            b = -b;
        }

        while (b > 0) {
    
    
            if ((b & 1) == 1) res *= x;
            x *= x;
            b >>= 1;
        }
        return res;
    }
}

猜你喜欢

转载自blog.csdn.net/qq_43349112/article/details/115335316
今日推荐