LeetCode-in-Java

50. Pow(x, n)

Medium

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

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 = 1/2² = 1/4 = 0.25

Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• -104 <= xn <= 104

Solution

public class Solution {
    public double myPow(double x, int n) {
        long nn = n;
        double res = 1;
        if (n < 0) {
            nn *= -1;
        }
        while (nn > 0) {
            if (nn % 2 == 1) {
                nn--;
                res *= x;
            } else {
                x *= x;
                nn /= 2;
            }
        }
        if (n < 0) {
            return 1.0 / res;
        }
        return res;
    }
}

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