50.pow-x-n.java

/*
 * @lc app=leetcode id=50 lang=java
 *
 * [50] Pow(x, n)
 *
 * https://leetcode.com/problems/powx-n/description/
 *
 * algorithms
 * Medium (27.87%)
 * Likes:    916
 * Dislikes: 2236
 * Total Accepted:    343.5K
 * Total Submissions: 1.2M
 * Testcase Example:  '2.00000\n10'
 *
 * 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]
 *
 *
 */
class Solution {
  public double myPow(double x, int n) {
    if (n == 0) {
      return 1;
    }
    double t = myPow(x, n/2);
    if (n % 2 != 0) {
      if (n < 0) {
        return t * t / x;
      } else {
        return t * t * x;
      }
    } else {
      return t * t;
    }
  }
}