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leetcode 50. pow(x, n)

admin 11月 13, 2020 0

50. Pow(x, n)

Difficulty:: Medium

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

Example 1:

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Input: 2.00000, 10
Output: 1024.00000

Example 2:

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Input: 2.10000, 3
Output: 9.26100

Example 3:

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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]

Solution

Language: Java

位运算

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class  {
    public double myPow(double x, int n) {
        if (x == 0) {
            return 0;
        }
        if (n == 0) {
            return 1;
        }
        long c = Math.abs((long) n);
        double result = 1;
        while (c > 0) {
            if ((c & 1) == 1) {
                result *= x;
            }
            x *= x;
            c >>= 1;
        }
        return n < 0 ? 1 / result : result;
    }
}

递归

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public class  {
    public double pow(double x, int n) {
        if(n == 0)
            return 1;
        if(n<0){
            n = -n;
            x = 1/x;
        }
        return (n%2 == 0) ? pow(x*x, n/2) : x*pow(x*x, n/2);
    }
}

递归麻烦版

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class  {
    public double myPow(double x, int n) {
        if (x == 0) {
            return 0d;
        }
        if (n == 0) {
            return 1d;
        }
        if (n < 0) {
            x = 1 / x;
            n = -n;
        }
        
        return pow(x, n);
        
    }
    
    private double pow(double x, int n) {
        if (n == 0) {
            return 1d;
        }
        if (n == 1) {
            return x;
        }
        if (n == 2) {
            return x * x;
        }
        double tmp = pow(x, n / 2);
        if (n % 2 == 0) {
            return tmp * tmp;
        } else {
            return tmp * tmp * x;
        }
    }
}
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