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:
Solutions#
Solution 1: Binary search#
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| 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
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| 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;
}
}
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| 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;
}
};
|
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| 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
}
|
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| 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;
}
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| 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
}
}
|
Solution 2: Math trick#
This is a math problem：
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| 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
|
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| class Solution:
def isPerfectSquare(self, num: int) -> bool:
i = 1
while num > 0:
num -= i
i += 2
return num == 0
|
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| class Solution {
public boolean isPerfectSquare(int num) {
for (int i = 1; num > 0; i += 2) {
num -= i;
}
return num == 0;
}
}
|
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| class Solution {
public:
bool isPerfectSquare(int num) {
for (int i = 1; num > 0; i += 2) num -= i;
return num == 0;
}
};
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| func isPerfectSquare(num int) bool {
for i := 1; num > 0; i += 2 {
num -= i
}
return num == 0
}
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| function isPerfectSquare(num: number): boolean {
let i = 1;
while (num > 0) {
num -= i;
i += 2;
}
return num === 0;
}
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| impl Solution {
pub fn is_perfect_square(mut num: i32) -> bool {
let mut i = 1;
while num > 0 {
num -= i;
i += 2;
}
num == 0
}
}
|
