# 633. Sum of Square Numbers

## Description

Given a non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c.

Example 1:

Input: c = 5
Output: true
Explanation: 1 * 1 + 2 * 2 = 5


Example 2:

Input: c = 3
Output: false


Constraints:

• 0 <= c <= 231 - 1

## Solutions

The picture above shows the relationship between a, b, and c. This question is actually looking up c in this table

From the upper right corner of the table, it is not difficult to find that it is similar to a binary search tree, so just start from the upper right corner and search according to the law of the binary search tree

• class Solution {
public boolean judgeSquareSum(int c) {
long a = 0, b = (long) Math.sqrt(c);
while (a <= b) {
long s = a * a + b * b;
if (s == c) {
return true;
}
if (s < c) {
++a;
} else {
--b;
}
}
return false;
}
}

• class Solution {
public:
bool judgeSquareSum(int c) {
long a = 0, b = (long) sqrt(c);
while (a <= b) {
long s = a * a + b * b;
if (s == c) return true;
if (s < c)
++a;
else
--b;
}
return false;
}
};

• class Solution:
def judgeSquareSum(self, c: int) -> bool:
a, b = 0, int(sqrt(c))
while a <= b:
s = a**2 + b**2
if s == c:
return True
if s < c:
a += 1
else:
b -= 1
return False


• func judgeSquareSum(c int) bool {
a, b := 0, int(math.Sqrt(float64(c)))
for a <= b {
s := a*a + b*b
if s == c {
return true
}
if s < c {
a++
} else {
b--
}
}
return false
}

• function judgeSquareSum(c: number): boolean {
let a = 0,
b = Math.floor(Math.sqrt(c));
while (a <= b) {
let sum = a ** 2 + b ** 2;
if (sum == c) return true;
if (sum < c) {
++a;
} else {
--b;
}
}
return false;
}


• use std::cmp::Ordering;
impl Solution {
pub fn judge_square_sum(c: i32) -> bool {
let c = c as i64;
let mut left = 0;
let mut right = (c as f64).sqrt() as i64;
while left <= right {
let num = left * left + right * right;
match num.cmp(&c) {
Ordering::Less => {
left += 1;
}
Ordering::Greater => {
right -= 1;
}
Ordering::Equal => {
return true;
}
}
}
false
}
}