Leetcode | Pow(x, n)

Implement pow(x, n).

快速求幂，注意n是负数的情况。

class Solution {
public:
    double pow(double x, int n) {
        double ans = 1.0;
        int symbol = 1;
        if (n < 0) {
            symbol = -1;
            n = -n;
        }
        for (; n > 0; n >>= 1) {
            if (n & 0x01) ans *= x;
            x *= x;
        }
        if (symbol < 0) return 1/ ans;
        else return ans;
    }
};