LeetCode – Refresh – Sqrt(x)

Newton method, need enought accuracy of tolerance to handle edge case INT_MAX and INT_MIN:

class Solution {
public:
    int sqrt(int x) {
        double first = x, second = 0;
        while (fabs(first - second) > 0.0000001) {
            second = first;
            first = (second + x/second)/2;
        }
        return int(first);
    }
};

 

Binary search, please remember check 0 and 1 for corner case:

class Solution {
public:
    int sqrt(int x) {
        if (x < 2) return x;
        int start = 0, end = x, mid = 0;
        while (end - start > 1) {
            mid = (end + start)/2;
            if (mid > 46340) end = mid;
            else if (mid*mid <= x) start = mid;
            else end = mid;
        }
        return start;
    }
};