Implement int sqrt(int x).

Compute and return the square root of x.

Have you met this question in a real interview? Yes

Example

sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

Challenge

O(log(x))

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class Solution {

public:

    /**

     * @param x: An integer

     * @return: The sqrt of x

     */

    int sqrt(int x) {

        // write your code here

        if (x < 1) {

            return 0;

        }

        long lo = 1;

        long hi = x / 2 + 1;

        while (lo < hi) {

            long mid = (lo + hi) / 2;

            long prod = mid * mid;

            if (prod <= x) {

                lo = mid + 1;

            } else {

                hi = mid;

            }

        }

        return lo - 1;

    }

};

还是依照upper_bound的思路