50. Pow(x, n)

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Problem

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

Constraints:

• -100.0 < x < 100.0
• $-2^{31} <= n <= 2^{31}-1$
• n is an integer.
• %-10^4 <= x^n <= 10^4%

Code

class Solution {
    public double myPow(double x, int n) {
        boolean isNegative = false;
        if (n < 0) {
            x = 1 / x;
            isNegative = true;
            // 避免溢出，当为MIN_VALUE时，可以防止溢出
            n = -(n + 1);
        }

        double ans = 1;
        double tmp = x;

        while (n != 0) {
            if (n % 2 == 1) {
                ans *= tmp;
            }
            tmp *= tmp;
            n /= 2;
        }

        if (isNegative) {
            ans *= x;
        }

        return ans;
    }
}