#Leetcode# 50. Pow(x, n)

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

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

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]

题解：快速幂

代码：

class Solution {
public:
    double myPow(double x, int n) {
        double ans = 1.0;
        for(int i = n; i != 0; i /= 2) {
            if(i % 2 != 0) ans *= x;
            x *= x;
        }
        if(n < 0) return (1 / ans);
        else return ans;
    }
};

#Leetcode# 50. Pow(x, n)

原文：https://www.cnblogs.com/zlrrrr/p/9966116.html