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package com.yangchd.leetcode.medium;

/**

* @author yangchd 2018/9/29.

*

* 50. Pow(x, n)

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

*

* 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 = 1/2^2 = 1/4 = 0.25

*

* Note:

* -100.0 < x < 100.0

* n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]

*/

public class PowXN {

class Solution {

public double myPow(double x, int n) {

if (n >= 0) {

if (n == 0) {

return 1;

}

if (n == 1) {

return x;

}

int count = 1;

double result = x;

while (count < n / 2) {

result = result * result;

count *= 2;

}

result = result * myPow(x, n - count);

return result;

}

if (n == Integer.MIN_VALUE) {

return 1 / myPow(x, Integer.MAX_VALUE) / x;

}

return 1 / myPow(x, Math.abs(n));

}

}

}