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 (30.39%)
 * Likes:    1757
 * Dislikes: 3277
 * Total Accepted:    529.4K
 * Total Submissions: 1.7M
 * Testcase Example:  '2.00000\n10'
 *
 * Implement pow(x, n), which calculates x raised to the power n (i.e. x^n).
 * 
 * 
 * Example 1:
 * 
 * 
 * Input: x = 2.00000, n = 10
 * Output: 1024.00000
 * 
 * 
 * Example 2:
 * 
 * 
 * Input: x = 2.10000, n = 3
 * Output: 9.26100
 * 
 * 
 * Example 3:
 * 
 * 
 * Input: x = 2.00000, n = -2
 * Output: 0.25000
 * Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25
 * 
 * 
 * 
 * Constraints:
 * 
 * 
 * -100.0 < x < 100.0
 * -2^31 <= n <= 2^31-1
 * -10^4 <= x^n <= 10^4
 * 
 * 
 */

// @lc code=start
class Solution {
    public double myPow(double x, int n) {
            double ans = 1;
            long absN = Math.abs((long)n);
            while(absN > 0) {
                if((absN&1)==1) ans *= x;
                absN >>= 1;
                x *= x;
            }
            return n < 0 ?  1/ans : ans;
        
    }
}
// @lc code=end