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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
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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 } } -
class Solution { public boolean judgeSquareSum(int c) { int n = (int) Math.sqrt(c); for (int i = 2; i <= n; ++i) { if (c % i == 0) { int exp = 0; while (c % i == 0) { c /= i; ++exp; } if (i % 4 == 3 && exp % 2 != 0) { return false; } } } return c % 4 != 3; } } -
class Solution { public: bool judgeSquareSum(int c) { int n = sqrt(c); for (int i = 2; i <= n; ++i) { if (c % i == 0) { int exp = 0; while (c % i == 0) { c /= i; ++exp; } if (i % 4 == 3 && exp % 2 != 0) { return false; } } } return c % 4 != 3; } }; -
class Solution: def judgeSquareSum(self, c: int) -> bool: for i in range(2, int(sqrt(c)) + 1): if c % i == 0: exp = 0 while c % i == 0: c //= i exp += 1 if i % 4 == 3 and exp % 2 != 0: return False return c % 4 != 3 -
func judgeSquareSum(c int) bool { n := int(math.Sqrt(float64(c))) for i := 2; i <= n; i++ { if c%i == 0 { exp := 0 for c%i == 0 { c /= i exp++ } if i%4 == 3 && exp%2 != 0 { return false } } } return c%4 != 3 } -
function judgeSquareSum(c: number): boolean { const n = Math.floor(Math.sqrt(c)); for (let i = 2; i <= n; ++i) { if (c % i === 0) { let exp = 0; while (c % i === 0) { c /= i; ++exp; } if (i % 4 === 3 && exp % 2 !== 0) { return false; } } } return c % 4 !== 3; }