LeetCode in Kotlin

69. Sqrt(x)

Easy

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

Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.

Note: You are not allowed to use any built-in exponent function or operator, such as pow(x, 0.5) or x ** 0.5.

Example 1:

Input: x = 4

Output: 2

Example 2:

Input: x = 8

Output: 2

Explanation: The square root of 8 is 2.82842…, and since the decimal part is truncated, 2 is returned.

Constraints:

• 0 <= x <= 231 - 1

Solution

class Solution {
    fun mySqrt(x: Int): Int {
        var start = 1
        var end = x / 2
        var sqrt = start + (end - start) / 2
        if (x == 0) {
            return 0
        }
        while (start <= end) {
            if (sqrt == x / sqrt) {
                return sqrt
            } else if (sqrt > x / sqrt) {
                end = sqrt - 1
            } else if (sqrt < x / sqrt) {
                start = sqrt + 1
            }
            sqrt = start + (end - start) / 2
        }
        return if (sqrt > x / sqrt) {
            sqrt - 1
        } else {
            sqrt
        }
    }
}

This site is open source. Improve this page.