# 367. Valid Perfect Square

## Description

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 product of some integer with itself.

You must not use any built-in library function, such as sqrt.

Example 1:

Input: num = 16
Output: true
Explanation: We return true because 4 * 4 = 16 and 4 is an integer.


Example 2:

Input: num = 14
Output: false
Explanation: We return false because 3.742 * 3.742 = 14 and 3.742 is not an integer.


Constraints:

• 1 <= num <= 231 - 1

## Solutions

Solution 1: Binary search

Solution 2: Math trick

This is a math problem：

1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = n²

• class Solution {
public boolean isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = (left + right) >>> 1;
if (mid * mid >= num) {
right = mid;
} else {
left = mid + 1;
}
}
return left * left == num;
}
}

• class Solution {
public:
bool isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = left + right >> 1;
if (mid * mid >= num)
right = mid;
else
left = mid + 1;
}
return left * left == num;
}
};

• class Solution:
def isPerfectSquare(self, num: int) -> bool:
left, right = 1, num
while left < right:
mid = (left + right) >> 1
if mid * mid >= num:
right = mid
else:
left = mid + 1
return left * left == num


• func isPerfectSquare(num int) bool {
left, right := 1, num
for left < right {
mid := (left + right) >> 1
if mid*mid >= num {
right = mid
} else {
left = mid + 1
}
}
return left*left == num
}

• function isPerfectSquare(num: number): boolean {
let left = 1;
let right = num >> 1;
while (left < right) {
const mid = (left + right) >>> 1;
if (mid * mid < num) {
left = mid + 1;
} else {
right = mid;
}
}
return left * left === num;
}


• use std::cmp::Ordering;
impl Solution {
pub fn is_perfect_square(num: i32) -> bool {
let num: i64 = num as i64;
let mut left = 1;
let mut right = num >> 1;
while left < right {
let mid = left + (right - left) / 2;
match (mid * mid).cmp(&num) {
Ordering::Less => {
left = mid + 1;
}
Ordering::Greater => {
right = mid - 1;
}
Ordering::Equal => {
return true;
}
}
}
left * left == num
}
}