leetcode - 69. Sqrt(x)

Description

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

Solution

Use binary search, since we want to find the rightest boundary so that the interval is all smaller than x, we discard the larger half.

Time complexity: o ( log ⁡ n ) o(\log n) o(logn)
Space complexity: o ( 1 ) o(1) o(1)

Code

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right = 0, x
        while left < right:
            mid = (left + right + 1) >> 1
            if mid * mid > x:
                right = mid - 1
            else:
                left = mid
        return (left + right + 1) >> 1

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