*SQRT (x)

Q:

Implement int sqrt (int x).

Compute and return the square root of X.

A:

Here are two methods of implementation: One is binary search, the other is Newton iterative method.

1.Two-point search

For a non-negative n, its square root is no greater than (n/2+1). In [0, n/2+1] This range can be a binary search, to find the square root of N.

1  Public classSolution {2      Public intMYSQRT (intx)3     {4        5      Longi = 0;6      Longj = x/2 + 1;7      while(I <=j)8     {9          LongMid = (i + j)/2;Ten          LongSq = Mid *mid; One         if(sq = = x)return(int) mid; A         Else if(Sq < x) i = mid + 1; -         Elsej = Mid-1; -     } the     return(int) J; -  -     } -}

Reference

Http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html

*SQRT (x)