69 Sqrt(x) - Easy

Problem:

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 ** 0.5 in python.

Example 1:

Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2.

Example 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.

Constraints:

• 0 <= x <= 231 - 1

Problem Analysis:

• use binary search to get O(log n)

Solutions:

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        # use binary search to achieve O(logn) instead of
        # brute force for loop: O(sqrt(n))

        l, r = 0, x
        res = 0

        while l <= r:
            # to avoid overflow, minus l not one
            m = l + (r - l) // 2
            if m*m > x:
                r = m-1
            elif m*m < x:
                l = m+1
                res = m
            else: 
                # exact match
                return m
        return res
        l, r = 0, x
        while l <= r:
            mid = l + (r-l)//2
            if mid * mid <= x < (mid+1)*(mid+1):
                return mid
            elif x < mid * mid:
                r = mid - 1
            else:
                l = mid + 1        ```

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