Skip to content

50. Pow(x, n)

Description

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

 

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

 

Constraints:

  • -100.0 < x < 100.0
  • -231 <= n <= 231-1
  • n is an integer.
  • Either x is not zero or n > 0.
  • -104 <= xn <= 104

Solutions

Solution 1: Mathematics (Fast Powering)

The core idea of the fast powering algorithm is to decompose the exponent $n$ into the sum of $1$s on several binary bits, and then transform the $n$th power of $x$ into the product of several powers of $x$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the exponent.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution:
    def myPow(self, x: float, n: int) -> float:
        def qpow(a: float, n: int) -> float:
            ans = 1
            while n:
                if n & 1:
                    ans *= a
                a *= a
                n >>= 1
            return ans

        return qpow(x, n) if n >= 0 else 1 / qpow(x, -n)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution {
    public double myPow(double x, int n) {
        return n >= 0 ? qpow(x, n) : 1 / qpow(x, -(long) n);
    }

    private double qpow(double a, long n) {
        double ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a;
            }
            a = a * a;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution {
public:
    double myPow(double x, int n) {
        auto qpow = [](double a, long long n) {
            double ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans *= a;
                }
                a *= a;
            }
            return ans;
        };
        return n >= 0 ? qpow(x, n) : 1 / qpow(x, -(long long) n);
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
func myPow(x float64, n int) float64 {
    qpow := func(a float64, n int) float64 {
        ans := 1.0
        for ; n > 0; n >>= 1 {
            if n&1 == 1 {
                ans *= a
            }
            a *= a
        }
        return ans
    }
    if n >= 0 {
        return qpow(x, n)
    }
    return 1 / qpow(x, -n)
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
function myPow(x: number, n: number): number {
    const qpow = (a: number, n: number): number => {
        let ans = 1;
        for (; n; n >>>= 1) {
            if (n & 1) {
                ans *= a;
            }
            a *= a;
        }
        return ans;
    };
    return n >= 0 ? qpow(x, n) : 1 / qpow(x, -n);
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
impl Solution {
    #[allow(dead_code)]
    pub fn my_pow(x: f64, n: i32) -> f64 {
        let mut x = x;
        let n = n as i64;
        if n >= 0 {
            Self::quick_pow(&mut x, n)
        } else {
            1.0 / Self::quick_pow(&mut x, -n)
        }
    }

    #[allow(dead_code)]
    fn quick_pow(x: &mut f64, mut n: i64) -> f64 {
        // `n` should greater or equal to zero
        let mut ret = 1.0;
        while n != 0 {
            if (n & 0x1) == 1 {
                ret *= *x;
            }
            *x *= *x;
            n >>= 1;
        }
        ret
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function (x, n) {
    const qpow = (a, n) => {
        let ans = 1;
        for (; n; n >>>= 1) {
            if (n & 1) {
                ans *= a;
            }
            a *= a;
        }
        return ans;
    };
    return n >= 0 ? qpow(x, n) : 1 / qpow(x, -n);
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
public class Solution {
    public double MyPow(double x, int n) {
        return n >= 0 ? qpow(x, n) : 1.0 / qpow(x, -(long)n);
    }

    private double qpow(double a, long n) {
        double ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans *= a;
            }
            a *= a;
        }
        return ans;
    }
}

Comments