#Leet Code# Sqrt

描述：log(n)

代码：

 class Solution:
     # @param x, an integer
     # @return an integer
     def getVal(self, begin, end, x):
         if end == begin :
             return begin
         if end == begin + 1:
             return begin

         while True:
             mid = (begin + end) / 2
             tmp = mid * mid 

             if tmp == x:
                 return mid
             elif tmp < x:
                 return self.getVal(mid, end, x)
             elif tmp > x:
                 return self.getVal(begin, mid, x)

     def sqrt(self, x):
         if x == 0: return 0

         val = 1
         while True:
             tmp  = val * val

             if tmp == x:
                 return val
             elif tmp > x:
                 return self.getVal(val/2, val, x)
             else:
                 val *= 2