#Leetcode# 50. Pow(x, n)
阿新 • • 發佈:2018-11-15
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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]
題解:快速冪
代碼:
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)