LeetCode_69. Sqrt(x)

69. Sqrt(x)

Easy

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.

package leetcode.easy;

public class SqrtX {
	@org.junit.Test
	public void test() {
		long x1 = 4;
		long x2 = 8;
		System.out.println(mySqrt(x1));
		System.out.println(mySqrt(x2));
	}

	public int mySqrt(long x) {
		for (long i = 0; i <= x; i++) {
			if (i * i == x) {
				return (int) i;
			} else if (i * i < x && (i + 1) * (i + 1) > x) {
				return (int) i;
			} else {
				continue;
			}
		}
		return 0;
	}
}