Sqrt(x)

Question: Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

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

Example 1:

Input: 4
Output: 2

Example 2:

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

You may view the full question here.

Approach 1: Let’s start with brute force —

//Approach 1
//Runtime: Time limit exceeded
//Memory usage: N/A
class Solution {
    public int mySqrt(int x) {
        for(int i = 1; i<=x; i++){
            if(i*i==x||((i+1)*(i+1))>x){
                return i;
            }
        }
        return 0;
    }
}

Approach 2: This is just for fun —

//Approach 2
//Runtime: 1ms
//Memory usage: 32.5MB
class Solution {
    public int mySqrt(int x) {
        return (int)(Math.sqrt(x));
    }
}

Clearly, we need another post to do justice. May be this might help.

Find more posts here.

Cheers & Chao!