Your program should print only one solution for each integer, even though there may be more than one solution in some cases.Try to make your program as efficient as possible. In particular, it is much more efficient to start with the largest possible perfect square and work down (as ...
02 第一种解法 暴力解法,定义一个long类型的变量,for循环判断其平方是否小于num,但是不要写循环体,这样省时,最后直接判断其平方是否等于num。 public boolean isPerfectSquare(intnum) {if(num<=1) {returnnum==1; } long i =1;for( ; i*i <num; i++);returni*i ==num; } 03 第二种解法 使用...
Credits: Special thanks to@elmirapfor adding this problem and creating all test cases. Subscribeto see which companies asked this question 解法一: 1publicclassSolution {2publicbooleanisPerfectSquare(intnum) {3if(num < 0)returnfalse;4if(num == 1)returntrue;5for(inti = 1; i<= num/i;i+...
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To solve this challenge, we need to write a Java program that takes an integer as input from the user and checks if it is a perfect square. A perfect square is a number that can be expressed as the square of another integer.
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Given a positive integer num, write a function which returns True if num is a perfect square else False. Example For example: Given num = 16 ReturnsTrue Solution class Solution { public boolean isPerfectSquare(int num) { if (num == 1) return true; ...
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