Pow(x, n) leetcode

Implement pow(x, n).

 

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利用依次消去二进制位上的1，来进行计算

double myPow(double x, int n) {
    double ans = 1;
    unsigned long long p;
    if (n < 0) {
        p = -n;
        x = 1 / x;
    }
    else {
        p = n;
    }
    while (p) {
        if (p & 1)
            ans *= x;
        x *= x;
        p >>= 1;
    }
    return ans;
}