LeetCode-in-Java

69. Sqrt(x)

Easy

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

• For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Example 1:

Input: x = 4

Output: 2

Explanation: The square root of 4 is 2, so we return 2.

Example 2:

Input: x = 8

Output: 2

Explanation: The square root of 8 is 2.82842…, and since we round it down to the nearest integer, 2 is returned.

Constraints:

• 0 <= x <= 231 - 1

Solution

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

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