LeetCode 367. Valid Perfect Square 题目解析 判断一个正数是否是完全平方数,不可使用sqrt函数。 解题思路 这是一道简单题,但是却挺有思考价值的。本题有很多种解法,思路各异,可以都看看。 无数种解法 解法一 数学里有一个概念叫做等差数列,完全平方数恰好是奇数等差数列之和。1+3+5+...+(2n-1)=(1+2n-1)n/2=n^2
【leetcode】367. Valid Perfect Square problem 367. Valid Perfect Square solution:二分法; classSolution {public:boolisPerfectSquare(intnum) {intleft =1, right = num;//longmid =0;while(left<=right) { mid= left +0.5*(right -left);longt = mid*mid;if(t < num) left = mid+1;elseif(...
Can you solve this real interview question? Valid Perfect Square - Given a positive integer num, return true if num is a perfect square or false otherwise. A perfect square is an integer that is the square of an integer. In other words, it is the produc
public boolean isPerfectSquare(int num) { int lo = 1; int hi = 46340; while (lo <= hi) { int mid = lo + (hi - lo) / 2; if (mid * mid < num) { lo = mid + 1; } else if (mid * mid > num) { hi = mid - 1; } else if (mid * mid == num){ return true; ...
367. Valid Perfect Square aliblielite 来自专栏 · leetcode_python_easyclass Solution: def isPerfectSquare(self, num): """ :type num: int :rtype: bool """ # 法一 # return (int(num**0.5)**2)==num # # 法二:牛顿迭代法 # r = num #...
valid perfect square(leetcode) valid perfect square 题目介绍 解题方法 最近在刷leetcode, 遇到一个很easy的问题,但是用蛮力的方法,根本解不开,想着应该是数学得某个知识,这个结介绍一下。 题目介绍 这个题目就是输入一个整数,来判断它是不是一个整数的平方,虽然可以用蛮力算法来解决,但是在提价的时候会遇到...
https://leetcode.com/problems/valid-perfect-square/ 题目: 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: ...
简介: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 True,否则返回 False。 Description 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. ...
LeetCode "Valid Perfect Square" Typical binary search.. but take care of data overflow if you are using C++ classSolution {public:boolisPerfectSquare(intnum) {if(num <0)returnfalse;if(num <2)returntrue;longlongi =1, j = num -1;while(i <=j)...
classSolution {public:boolisPerfectSquare(intnum) {longx =num;while(x * x >num) { x= (x + num / x) /2; }returnx * x ==num; } }; 本文转自博客园Grandyang的博客,原文链接:检验完全平方数[LeetCode] Valid Perfect Square,如需转载请自行联系原博主。