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]
AC code:
Recursive:
class Solution { public: double myPow(double x, int n) { if (n >= 0) return pow(x, n); else return 1.0 / pow(x, n); } double pow(double x, int n) { if (n == 0) { return 1; } double y = pow(x, n / 2); if (n % 2 == 0) { return y * y; } else { return y * y * x; } } };
Runtime: 4 ms, faster than 99.40% of C++ online submissions for Pow(x, n).
Iteration:
class Solution { public: double myPow(double x, int n) { long y = abs((long)n); double res = 1; while (y > 0) { if (y % 2 != 0) { res *= x; } x *= x; y /= 2; } if (n < 0) { res = 1.0 / res; } return res; } };
Runtime: 4 ms, faster than 99.40% of C++ online submissions for Pow(x, n).