原题: 367. Valid Perfect Square 读题: 求一个整数是否为完全平方数,如1,4,9,16,……就是完全平方数,这题主要是运算效率问题 求解方法1:812ms class Solution { public: bool isPerfectSquare(int num) { if(num < 0) return false; int i = 0; //这里要加1,不然num = 1时会出错 for(;i< num/2 + 1;i++) { if(i*i == num) {…

Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Returns: True Example 2: Input: 14 Returns: False Credits:Speci…

Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Returns: True Example 2: Input: 14 Returns: False 本题可以用binary s…

题目描述: Given a positive integer num, write a function which returns True if num is a perfect square else False. 解题分析: 这种找数字的题一般都用类似与二分查找的算法.需要注意的是比较平方和时考虑到integer溢出的情况.所以这个结果是要用Long类型保存.由此到来的改变是判断相等时要用“equals()”方法,而不是“==”. 实现代码: public class Solution…

Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Returns: True Example 2: Input: 14 Returns: False 题意: 验证完全平方数 思…

［抄题］: Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Returns: True Example 2: Input: 14 Returns: False [暴力解法]:…

Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Output: true Example 2: Input: 14 Output: false class Solution…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:完全平方式性质 方法二:暴力求解 方法三:二分查找 方法四:牛顿法 日期 题目地址:https://leetcode.com/problems/valid-perfect-square/description/ 题目描述 Given a positive integer num, write a function which returns…

题目描述:给出一个正整数,不使用内置函数,如sqrt(),判断这个数是不是一个数的平方. 思路:直接使用二分法,貌似没啥好说的.代码如下: class Solution(object): def isPerfectSquare(self, num): """ :type num: int :rtype: bool """ left, right = 0, num while left <= right: mid = (left+right)…

class Solution { public: bool isPerfectSquare(int num) { /* //方法一:蜜汁超时…… if (num < 0) return false; if (num == 1) return true; for (int i=1;i*i<=num;i++){ if (i*i==num) return true; } return false; }*/ /* //方法二:是对的! if(num < 0) return false; if(n…