LeetCode - Square root of #69 x

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Difficulty Level: Easy

1. Description

Given a non-negative integer x, compute and return xthe arithmetic square root of .

Since the return type is an integer, only the integer part of the result is retained, and the fractional part will be rounded .

Note: Any built-in exponential functions and operators such as pow(x, 0.5)or x ** 0.5.

2. Example

Example 1

输入：x = 4
输出：2
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Example 2

输入：x = 8
输出：2
解释：8 的算术平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。
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Restrictions:

0 <= x <= 2^31 - 1

3. Answers

class Sqrtx {
    func mySqrt(_ x: Int) -> Int {
        guard x >= 0 else {
            return 0
        }
        
        var left = 0, right = x / 2 + 1
        
        while left <= right {
            let mid = (right - left) / 2 + left
            
            if mid * mid == x {
                return mid
            } else if mid * mid < x {
                left = mid + 1
            } else {
                right = mid - 1
            }
        }
        
        return right
    }
}
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Main idea: binary search, should x / 2 + 1start with , so its square is x + x ^ 2 / 4 + 1, it must be greater than x.
Time Complexity: O(logn)
Space Complexity: O(1)

Note: Please use (right-left)/2+left to get middle in case of integer overflow

Repository for the algorithm solution: LeetCode-Swift

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