[Leetcode 85] 69 Sqrt(x)

Problem:

Implement int sqrt(int x).

Compute and return the square root of x.

 

Analysis:

If we simply use the naive algorithm to go through all the numbers from 0 until the square root, it will exceed the time limit. The hero is binary search.

We want such a number between 0 and n inclusively that x*x <= n and  (x+1)*(x+1) > n. Not exactly x*x = n because its the integer square root. With this exit condition, we can easily use binary search to get the final result.

 

Code:

class Solution {
public:
    int sqrt(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int s = 0, e = x;
        
        while (s <= e) {
            long long mid = (s + e) / 2, sq = mid * mid;
            
            if (sq <= x && (mid+1) * (mid+1) > x)
                return (int) mid;
            else if (sq > x)
                e = mid - 1;
            else
                s = mid + 1;
        }
        
        return -1;
    }
};