Given a non-negative integer
x, return the square root of
x rounded down to the nearest integer. The returned integer should be non-negative as well.
You must not use any built-in exponent function or operator.
- For example, do not use
pow(x, 0.5)in c++ or
x ** 0.5in python.
Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2.
Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.
0 <= x <= 231 - 1
class Solution(object): def mySqrt(self, x): """ :type x: int :rtype: int """ # right is set to x + 1 to deal with special cases like x = 0, x = 1 left, right = 0, x + 1 while left < right: mid = left + (right - left) // 2 # Square root means x*x = y if mid * mid > x: right = mid else: left = mid + 1 # because of our loop we break when we encounter a mid which when squared is more than our target # therefore our answer is the previous integer return left - 1