Sqrt(x).md

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4 Output: 2 Example 2:

Input: 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.

My code:

class Solution:
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        if x==0:
            return 0
        result=self.Sqrt(x,0,x)
        return int(result)
    def Sqrt(self,x,i,j):
        if j-i<=2:
            check_index=i
            for k in range(i,j+1):
                if k**2<=x:
                    check_index=k
            assert (check_index+1)**2>x
            return check_index
        mid=int((i+j)/2)
        if mid**2>x:
            return self.Sqrt(x,i,mid)
        else:
            return self.Sqrt(x,mid,j)

Solution: https://leetcode.com/problems/sqrtx/