本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试。 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...
解法一: 1publicclassSolution {2publicbooleanisPerfectSquare(intnum) {3if(num < 0)returnfalse;4if(num == 1)returntrue;5for(inti = 1; i<= num/i;i++){6if(i*i == num)returntrue;7}8returnfalse;9}10} 解法2: 二分查找法 (java 没有ac 仅供参考) 1publicclassSolution {2publicboolean...
public class Solution { public boolean isPerfectSquare(int num) { for(int i = 1; num > 0; i += 2){ num -= i; } return num == 0; } } 1. 2. 3. 4. 5. 6. 7. 8. python版本: class Solution(object): def isPerfectSquare(self, num): """ :type num: int :rtype: bool...
int n=0 Scanner in=new Scanner(System.in) doSystem.out.print("Enter a positive integer (-1 to quit): ") n=in.nextInt() if(n 相关知识点: 试题来源: 解析 把findSum中的 if(findSum(n-largestSquare(n),numOfBreak+1))改成 if(findSum(n-i,numOfBreak+1))...
Problem 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; ...
javamathoptimizationperfect-square 1605 我正在寻找最快的方法来确定一个 long 值是否是一个完全平方数(即其平方根是另一个整数): 我已经用内置的Math.sqrt()函数轻松完成了它,但我想知道是否有一种方法可以通过限制自己到整数域来更快地完成。 维护查找表是不切实际的(因为大约有231.5个整数的平方小于263)。
Perfect Squares Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 460 Accepted Submission(s): 256 Problem Description A number x is called a perfect square if there exists an integer b ...
A number x is said to be a perfect square if there exists an integer y such that x = y2. Input The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array. The second line contains n integers a1, a2, ..., an (...
Program to check if given number is perfect square in Kotlin packagecom.includehelpimport java.util.*//import kotlin.math.floor//import kotlin.math.sqrt//function to check perfect Square numberfuncheckPerfectSquare(number:Double):Boolean{//Square Root of Numbervalsq= Math.sqrt(number)//Floor va...
今天的题:链接:https://leetcode-cn.com/problems/valid-perfect-square给定一个正整数num,编写一个函数,如果num是一个完全平方数,则返回...num/2+1分别平方与num作比较,若存在相等的情况,则返回true,若循环结束,返回false.因为num/2+1必定会大于sqrt(num),即大于num的开方,即使这样能实现要求,但最后的运行...