LeetCode – Pow(x, n)

 

Problem:

Implement pow(x, n).

This is a great example to illustrate how to solve a problem during a technical interview. The first and second solution exceeds time limit; the third and fourth are accepted.

Java Solution

public double myPow(double x, int n){
    if(n==0)
        return 1;
 
    if(n<0){
        return 1/helper(x, -n);
    }
 
    double v = helper(x, n/2);
 
    if(n%2==0){
       return v*v;
    }else{
       return v*v*x;
    }    
}

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Category >> Algorithms  
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  • Alik Elzin

    Where’s the helper() method?

  • Nikhil Bagde

    Doesn’t work for base = 10 and power = 3
    Gives result = 100.

  • Bruce

    this is a very good idea, I polish it a little as following in Java
    // nonrecursive implementation

    // its not the fast one, only runtime beats 29.88% of java submissions on 2/1/2016.

    static double power2(double x, int n) {

    if (x == 1) {

    return 1;

    }

    if (x == -1) {

    return n % 2 == 0 ? 1 : -1;

    }

    boolean negative = false;

    if (n >= 1;

    }

    return negative ? 1 / result : result;

    }

  • Anushree Acharjee

    the Best Solution given here has Overflow error. e,.g pow(71045970,41535484)

  • hello

    what if power < 0?

  • carllee1991

    best one:

    static double pow(double x, int n) {
    if (n == 0) return 1;
    double half = pow(x, n / 2);
    if (n % 2 == 0) return half * half;
    else if (n > 0) return half * half * x;
    else return half * half / x;
    }

  • There is another simple solution for the problem

    public int pow(int x, int n) {
    int result = 1;

    while (n > 0) {
    if (n % 2 != 0) {
    result = result * x;
    }

    x = x * x;
    n = n / 2;
    }

    return result;
    }

  • Amir

    Iterative solution:

  • Aphiwad Chhoeun

    My solution :

    public double pow(int x, int n) {
    if (x == 0) return 0;
    if (n == 0) return 1;
    if(n<0) return 1/pow(x, -n);
    return x * pow(x, n – 1);
    }

  • Sébastien Dijols

    Thanks for solutions regarding this problem.

    Just a little remark regarding your third solution. The main issue is that it’s not actually using the advantage of that recursive definition. You shouldn’t actually call pow(x, half) * pow(x, half) as that’s doing the recursion twice to return the same result. Just store the value in a variable, and you should have log(n) computations.
    ie:
    int halfPower = pow(x, half);
    return halfPower * halfPower * pow(x, remainder);

  • codeaperan

    I have found a more concise solution from https://github.com/yuanx/leetcode/blob/master/Pow.java

    public double pow(double x, int n) {
    if (n == 0) return 1;
    if (n == 2) return x * x;
    if (n % 2 == 0) return pow(pow(x, n / 2), 2);
    else if (n > 0) return x * pow(pow(x, n / 2), 2);
    else return 1 / x * pow(pow(x, n / 2), 2);
    }

  • Dan

    Need to handle: n=-2147483648.

  • holey

    cool solution 4.