50. Pow(x, n)（递归）

50. Pow(x, n)

Medium

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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]

 

 

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0: return 1
        if n < 0: 
            x = 1/x 
            n = -n
        res = 1
        while n > 0:
            if n & 1: res *= x
            x *= x
            n = n>>1
        return res

 

 

 

class Solution {
public:
    double myPow_help(double x, long n) {
        if (n==0) return 1;
        if(n==1) return x;    
        double tt = myPow_help(x,n/2);
        return (n%2==1)?tt*tt*x:tt*tt;
    }
    double myPow(double x, int n) {
        return n>=0? myPow_help(x,n): 1/myPow_help(x,-long(n));
    
    }
};

 

 

 

class Solution {
public:
    double myPow(double x, int n) {
        if (n==-2147483648) return Pow(x,n+1)/x;
        else return Pow(x,n);
        
    }
    double Pow(double x, int n) {
        if (x==1.0 || x==0.0) return x;
        if(n > 0) {
            double res = myPow(x,n/2);
            if (n%2==0) return res*res;
            else return res*res*x;
        } else if (n < 0) {
            double res = myPow(x,-n/2);
            if(((-n)%2)==0) return 1/(res*res);
            else return 1/(res*res*x);
        } else {
            return 1;
        }
    }
};