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.
1 # @param {float} x 2 # @param {integer} n 3 # @return {float} 4 def myPow(self, x, n): 5 pz=1 6 if n<0: 7 pz=-1 8 n=abs(n) 9 r=1 10 base=x 11 while n!=0: 12 if n%2==1: 13 r*=base 14 base*=base 15 n/=2 16 return r if pz==1 else 1.0/r
Note that the sign of n should be handled.