69. Sqrt(x).rs

impl Solution {
    pub fn my_sqrt(x: i32) -> i32 {
        let mut i = x as f64;
        while (i*i - x as f64).abs() > 0.75_f64 {
            i = (i+(x as f64/i))/2_f64;
        }
        return i.floor() as i32;
    }
}


// The overall time complexity is O(Log(n)), where n represents the input integer x.
// 
// The process I went through here is that I knew I could use Newton's method to approximate the square root of x, making
// the overall time complexity O(Log(n)). Here is the breakdown of my code:
//     1. I start my function by initializing the integer i equal to the target value x.
//     2. Then I open a while-loop which keeps on iterating through Newton's method until our estimate i is a close enough
//        estimate for the square root of x (within 0.75).
//     3. Then I simply return the floored integer from the estimate i.