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

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 = 1/2² = 1/4 = 0.25

Note:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, within the range [−2³¹, 2³¹ − 1]

AC代码

class Solution {
public:
    double myPow(double x, int n) {
        if (x == 1 || x == 0) return x;
        if (n == 0) return 1;
        double ans = 1.0;
        for (int i = n; i != 0; i /= 2) {
            if (i % 2 != 0) ans *= x;
            x *= x;
        }
        return n > 0 ? ans : 1 / ans;
    }
};

总结

还有快速幂的相关解法，有时间再学