50. Pow(x, n)

发布时间：2020-12-14 04:18:37 所属栏目：大数据来源：网络整理

导读：Implement?pow( x ,? n ),which calculates? x ?raised to the power? n ?(xn). Example 1: Input: 2.00000,10Output: 1024.00000 Example 2: Input: 2.10000,3Output: 9.26100 Example 3: Input: 2.00000,-2Output: 0.25000Explanation: 2-2 = 1/22 = 1/4 =

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 code:

Recursive:

class Solution {
public:
    double myPow(double x,int n) {
        if (n >= 0) return pow(x,n);
        else return 1.0 / pow(x,n);
    }
    
    double pow(double x,int n) {
        if (n == 0) {
            return 1;
        }
        double y = pow(x,n / 2);
        if (n % 2 == 0) {
            return y * y;
        } else {
            return y * y * x;
        }
    }
};

Runtime:? 4 ms,faster than?99.40%?of?C++?online submissions for?Pow(x,n).

Iteration:

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

};

Runtime:?4 ms,n).

（编辑：李大同）

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