LeetCode_69. Sqrt(x)

发布时间：2020-12-14 00:22:25 所属栏目：Linux 来源：网络整理

导读：? 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

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;
	}
}

（编辑：李大同）

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