JavaScript 中测试给定参数是否为平方数的最佳方法是什么？
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What's the best way in JavaScript to test if a given parameter is a square number?
提问by Mike. D
I created a function that will test to see if a given parameter is a square number.
我创建了一个函数来测试给定的参数是否是一个平方数。
Read about square numbers here: https://en.wikipedia.org/?title=Square_number
在此处阅读平方数：https: //en.wikipedia.org/?title=Square_number
If the number is a square number, it returns trueand otherwise false. Negative numbers also return false.
如果数字是平方数，则返回true，否则返回false。负数也返回false。
Examples:
例子：
isSquare(-12) // => false
isSquare( 5) // => false
isSquare( 9) // => true
isSquare(25) // => true
isSquare(27) // => false
Right now, I am using this method: http://jsfiddle.net/marcusdei/ujtc82dq/5/
现在，我正在使用这种方法：http: //jsfiddle.net/marcusdei/ujtc82dq/5/
But, is there a shorter more cleaner way to get the job done?
但是，是否有更短更清洁的方法来完成工作？
回答by Tushar
回答by Stephane Moreau
回答by ASHISH RANJAN
//1st
var isPerfectSquare = function(num) {
return Math.sqrt(num) % 1 === 0;
}
//2nd: loop through all the number from 1 to num
var isPerfectSquare = function(num) {
for(let i=1; i <= num ; i++){
let d = i * i;
if(d === num){
return true
}
}
}
// Optimize solution: Binary Search
var isPerfectSquare = function(num) {
if(num ==1)return true
let left = 2;
let right = Math.floor(num/2);
while(left <= right){
let middle = Math.floor((left + right)/2)
let sqr = middle * middle;
if(sqr == num){
return true
}else{
if(sqr > num){
right = middle -1
}else{
left = middle + 1
}
}
}
return false
};回答by Jeff Lowery
It's a bit trickier if you're using the new BigInt in JavaScript:
如果您在 JavaScript 中使用新的 BigInt，这会有点棘手：
// integer square root function (stolen from the interwebs)
function sqrt(n) {
let a = 1n;
let b = (n >> 5n) + 8n;
while (b >= a) {
let mid = (a + b) >> 1n;
if (mid * mid > n) {
b = mid -= 1n;
} else {
a = mid += 1n;
}
}
return a -= 1n;
}
sqrt(25n) === 5n
sqrt(26n) === 5n
...
sqrt(35n) === 5n
The best and fastest way I've found (so far) to determine if n is a square is:
我发现（到目前为止）确定 n 是否为正方形的最好和最快的方法是：
function isSquare(n) {
return n%sqrt(n) === 0n
}
But there's gotta be a faster way for BigInt operations.
但是必须有一种更快的方式来进行 BigInt 操作。
回答by AnonymousContribute
Isn't this (Math.sqrt(number) % 1 === 0) basically enough? it just checks if the sqrt of the number is a whole number, if so, then it's a perfect square.
这不是 (Math.sqrt(number) % 1 === 0) 基本上够了吗？它只是检查数字的 sqrt 是否是整数，如果是，则它是一个完美的正方形。
Obviously, depending on what you want to do with that information, it may require extra code.
显然，根据您想对这些信息做什么，它可能需要额外的代码。
回答by A.McLoof
I think that this one is a shorter and a cleaner option:
我认为这是一个更短更干净的选择：
var isSquare = function(n) {
return Number.isInteger(Math.sqrt(n));
};
isSquare(25); //true
for even shorter and cleaner than that you could use:
比你可以使用的更短更干净：
var isSquare = n => Number.isInteger(Math.sqrt(n));
isSquare(25);//true
