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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

  • 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
    }
}

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