检查给定的数字N是否是一个完美的平方。如果是,则返回它是一个完美正方形的数字,否则打印-1。
例子:
Input: N = 4900
Output 70
Explanation:
4900 is a perfect square number of 70 because 70 * 70 = 4900
Input: N = 81
Output: 9
Explanation:
81 is a perfect square number of 9 because 9 * 9 = 81
方法:为解决上述问题,我们将使用二进制搜索算法。
- 从起始值和最后一个值中找到mid元素,然后将mid(mid * mid)的平方值与N进行比较。
- 如果相等,则返回中值,否则检查square(mid * mid)是否大于N,然后以相同的起始值但最后更改为mid-1值的方式进行递归调用,并且square(mid * mid)小于然后,N的递归调用具有相同的最后一个值,但更改了起始值。
- 如果N不是平方根,则返回-1。
下面是上述方法的实现:
C++
// C++ program to check if a
// given number is Perfect
// square using Binary Search
#include
using namespace std;
// function to check for
// perfect square number
int checkPerfectSquare(
long int N,
long int start,
long int last)
{
// Find the mid value
// from start and last
long int mid = (start + last) / 2;
if (start > last) {
return -1;
}
// check if we got the number which
// is square root of the perfect
// square number N
if (mid * mid == N) {
return mid;
}
// if the square(mid) is greater than N
// it means only lower values then mid
// will be possibly the square root of N
else if (mid * mid > N) {
return checkPerfectSquare(
N, start, mid - 1);
}
// if the square(mid) is less than N
// it means only higher values then mid
// will be possibly the square root of N
else {
return checkPerfectSquare(
N, mid + 1, last);
}
}
// Driver code
int main()
{
long int N = 65;
cout << checkPerfectSquare(N, 1, N);
return 0;
}
Java
// Java program to check if a
// given number is Perfect
// square using Binary Search
import java.util.*;
class GFG {
// Function to check for
// perfect square number
static int checkPerfectSquare(long N,
long start,
long last)
{
// Find the mid value
// from start and last
long mid = (start + last) / 2;
if (start > last)
{
return -1;
}
// Check if we got the number which
// is square root of the perfect
// square number N
if (mid * mid == N)
{
return (int)mid;
}
// If the square(mid) is greater than N
// it means only lower values then mid
// will be possibly the square root of N
else if (mid * mid > N)
{
return checkPerfectSquare(N, start,
mid - 1);
}
// If the square(mid) is less than N
// it means only higher values then mid
// will be possibly the square root of N
else
{
return checkPerfectSquare(N, mid + 1,
last);
}
}
// Driver code
public static void main(String[] args)
{
long N = 65;
System.out.println(checkPerfectSquare(N, 1, N));
}
}
// This code is contributed by offbeat
Python3
# Python3 program to check if a
# given number is perfect
# square using Binary Search
# Function to check for
# perfect square number
def checkPerfectSquare(N, start, last):
# Find the mid value
# from start and last
mid = int((start + last) / 2)
if (start > last):
return -1
# Check if we got the number which
# is square root of the perfect
# square number N
if (mid * mid == N):
return mid
# If the square(mid) is greater than N
# it means only lower values then mid
# will be possibly the square root of N
elif (mid * mid > N):
return checkPerfectSquare(N, start,
mid - 1)
# If the square(mid) is less than N
# it means only higher values then mid
# will be possibly the square root of N
else:
return checkPerfectSquare(N, mid + 1,
last)
# Driver code
N = 65
print (checkPerfectSquare(N, 1, N))
# This code is contributed by PratikBasu
C#
// C# program to check if a
// given number is Perfect
// square using Binary Search
using System;
class GFG{
// Function to check for
// perfect square number
public static int checkPerfectSquare(int N,
int start,
int last)
{
// Find the mid value
// from start and last
int mid = (start + last) / 2;
if (start > last)
{
return -1;
}
// Check if we got the number which
// is square root of the perfect
// square number N
if (mid * mid == N)
{
return mid;
}
// If the square(mid) is greater than N
// it means only lower values then mid
// will be possibly the square root of N
else if (mid * mid > N)
{
return checkPerfectSquare(N, start,
mid - 1);
}
// If the square(mid) is less than N
// it means only higher values then mid
// will be possibly the square root of N
else
{
return checkPerfectSquare(N, mid + 1,
last);
}
}
// Driver code
public static int Main()
{
int N = 65;
Console.Write(checkPerfectSquare(N, 1, N));
return 0;
}
}
// This code is contributed by sayesha
Javascript
输出:
-1
时间复杂度: O(登录)