#Leetcode# 50. Pow(x, n)

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

 

Implement pow(xn), 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;
    }
};

  

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