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.

04/15/2019 UPdate: 利用binary search，找last index that i * i <= x.

Time: O(lg n)   , Space: O(1)

Code

class Solution:

    def sqrt(self, x):

        if x < 2: return x

        l, r = 1, x  # r * r > x for sure

        while l + 1 < r:

            mid = l + (r - l)//2

            value = mid * mid

            if value > x:

                r = mid

            elif value < x:

                l = mid

            else:

                return mid

        return l

class Solution:

    def sqrt(self, num):

        ans = num

        while ans*ans > num:

            ans = (ans + num//ans)//2

        return ans