LeetCode#50 Pow(x, n)

Just...

Implement pow(x, n).

Solution:

1）Naive solution：multiply x by itself for n-1 times. (Or simply reyurn 1 if n==0).

This takes O(n) time. When n is big enough, it's relatively slow.

2）Fast Power.

Take pow(3, 11) as an example. Here's what we're gonna do:

represent 11 as a binary number, which is bn=1011. Assign x to a new variable base=x, and set result as 1.

iterate digits in bn from rear to head,

    1._if the current digit is 1, result is multiplied by base;

    2._if the current digit is 0, result will remain unchanged;

    3._alter base: base=base * base.

The reversed iteration of binary digits can alse be replaced by a series of dived-and-mod operations on n.

Mathematics behind it:

Each n has a baniry version. And we know: pow(a, m+n)=pow(a, m)*pow(a, n).

Thus pow(3, 11)=pow(3, 1)*pow(3, 2)*pow(3, 8).

When iterating over these binary digits backwards, 1 means the corresponding power of base is multiplied to

the final result, and 0 means not.

    # @param {float} x
    # @param {integer} n
    # @return {float}
    def myPow(self, x, n):
        pz=1
        if n<0:
            pz=-1
            n=abs(n)
        r=1
        base=x
        while n!=0:
            if n%2==1:
                r*=base
            base*=base
            n/=2
        return r if pz==1 else 1.0/r

 Note that the sign of n should be handled.