int raise(num, power){

if(power==0) return 1;

if(power==1) return num;

return(raise(num, power-1)*num);

}

11 Answers

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13

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int raise(num, power){

if(power==0) return 1;

if(power==1) return num;

return(raise(num, power-1)*num);

}

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6

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double Power(int x, int y)

{

double ret = 1;

double power = x;

while (y > 0)

{

if (y & 1)

{

ret *= power;

}

power *= power;

y >>= 1;

}

return ret;

}

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1

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In Ruby:

def power(base, power)

product = 1

power.times do

product *= base

end

product

end

puts "2^10 = 1024 = #{power(2,10)}"

puts "2^0 = 1 = #{power(2,0)}"

puts "2^1 = 2 = #{power(2,1)}"

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2

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If I were an interviewer, I would ask the Aug 29, 2010 poster why he used bitwise operators, and whether he would deploy that code in a production environment, or if he merely wanted to demonstrate, for purposes of the interview, that he understands bitwise operations.

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0

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Because it uses dynamic programming and is lots more efficient than your algorithm.

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0

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If the power is not integer, use ln and Taylor series

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1

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If I'm the interviewer, none of above answers is acceptable.

What if y < 0? what if y < 0 and x == 0?

I'm seeing an endless recursion that will eventually overflow the stack, and the none-recursive one just simply returns 1.

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0

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There is a way to do this in a logN way rather than N.

function power(x, n)

{

if n == 1

return x;

// Even numbers

else if (n%2 == 0)

return square( power (x, n/2));

// Odd numbers

else

return power(x, n-1);

}

This is from Programming pearls.. interesting way.

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7

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small mistake

function power(x, n)

{

if n == 1

return x;

// Even numbers

else if (n%2 == 0)

return square( power (x, n/2));

// Odd numbers

else

return power(x, n-1) * x;

}

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0

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# Solution for x ^ n with negative values of n as well.

def square(x):

return x * x

def power(x, n):

if x in (0, 1):

return x

if n == 0:

return 1

if n < 0:

x = 1.0 / x

n = abs(n)

# Even number

if n % 2 == 0:

return square(power(x, n/2))

# Odd number

else:

return x * power(x, n - 1)

print ("0 ^ 0 = " + str(power(0, 0)))

print ("0 ^ 1 = " + str(power(0, 1)))

print ("10 ^ 0 = " + str(power(10, 0)))

print ("2 ^ 2 = " + str(power(2, 2)))

print ("2 ^ 3 = " + str(power(2, 3)))

print ("3 ^ 3 = " + str(power(3, 3)))

print ("2 ^ 8 = " + str(power(2, 8)))

print ("2 ^ -1 = " + str(power(2, -1)))

print ("2 ^ -2 = " + str(power(2, -2)))

print ("2 ^ -8 = " + str(power(2, -8)))

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pretty trivial...