版权声明:Iori 的技术分享,所有内容均为本人原创,引用请注明出处,谢谢合作! https://blog.csdn.net/qq_31116753/article/details/82889107
问题
实现 int sqrt(int x) 函数。
计算并返回 x 的平方根,其中 x 是非负整数。
由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去。
输入: 4
输出: 2
输入: 8
输出: 2
说明: 8 的平方根是 2.82842..., 由于返回类型是整数,小数部分将被舍去。
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.
Input: 4
Output: 2
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
示例
public class Program {
public static void Main(string[] args) {
var n = 8;
var res = MySqrt(n);
Console.WriteLine(res);
n = 168;
res = MySqrt2(n);
Console.WriteLine(res);
n = 69;
res = MySqrt3(n);
Console.WriteLine(res);
Console.ReadKey();
}
private static int MySqrt(int x) {
//耍赖
return (int)(Math.Sqrt(x));
}
private static int MySqrt2(int x) {
//二分逼近,最后收敛于sqrt(x)
long res = x;
while(res * res > x) {
res = (res + x / res) / 2;
}
return (int)res;
}
private static int MySqrt3(int x) {
//魔数0x5f3759df,这个解法自行百度
//另外,这个解法LeetCode未AC,因为不支持unsafe
var i = 0L;
var num1 = 0F;
var num2 = 0F;
const float f = 1.5F;
num1 = x * 0.5F;
num2 = x;
unsafe {
i = *(long*)&num2;
i = 0x5f3759df - (i >> 1);
num2 = *(float*)&i;
}
num2 = num2 * (f - (num1 * num2 * num2));
num2 = num2 * (f - (num1 * num2 * num2));
return (int)(x * num2);
}
}
以上给出3种算法实现,以下是这个案例的输出结果:
2
12
8
分析:
MySqrt 的时间复杂度依赖于运行库的实现,MySqrt2 的时间复杂度为: ,MySqrt3 的时间复杂度为: 。