上篇文章中一道数学问题 - 自除数，今天我们接着分析 LeetCode 中的另一道数学题吧~

今天要给大家分析的面试题是 LeetCode 上第 633 号问题，

Leetcode 633 - 平方数之和

https://leetcode.com/problems/sum-of-square-numbers/

题目描述

给定一个非负整数 c ，你要判断是否存在两个整数 a和 b，使得 \(a^2 + b^2 = c\)。

示例1:

输入: 5
输出: True
解释: 1 * 1 + 2 * 2 = 5

示例2:

输入: 3
输出: False

Input:

5
2
100

Expected answer:

true
true
true

• 题目难度: 简单

• 贡献者： Stomach_ache

相关话题

• 数学

  https://leetcode-cn.com/tag/math

相似题目

• 有效的完全平方数
  https://leetcode-cn.com/problems/valid-perfect-square

解题思路:

做一次循环，用目标和减去循环变量的平方，如果剩下的部分依然是完全平方的情形存在，就返回true；否则返回false。

假定 $i \leq a \leq b $，根据数据的对称性，循环变量 i 只需取到 $i^2 \cdot 2 \leq c $ 即可覆盖所有情形.

已AC代码:

最初版本:

public class Solution
{
    public bool JudgeSquareSum(int c)
    {           
        for (int i = 0; c - 2 * i * i >= 0; i++)
        {
            double diff = c - i*i;
            if ((int)(Math.Ceiling(Math.Sqrt(diff))) == (int)(Math.Floor(Math.Sqrt(diff))))  // 若向上取整=向下取整，则该数开方后是整数
                return true;
        }

        return false;
    }
}

Rank:

执行用时: 56 ms, 在所有 csharp 提交中击败了68.18%的用户.

优化1:

public class Solution
{
    public bool JudgeSquareSum(int c)
    {           
        for (int i = 0; c - 2 * i * i >= 0; i++)
        {
            int diff = c - i*i;
            if (IsPerfectSquare(diff))
                return true;
        }

        return false;
    }
    private bool IsPerfectSquare(int num)
    {
        double sq1 = Math.Sqrt(num);
        int sq2 = (int)Math.Sqrt(num);
        if (Math.Abs(sq1 - (double)sq2) < 10e-10)
            return true;
        return false;
    }
}

Rank:

执行用时: 52 ms, 在所有 csharp 提交中击败了90.91%的用户.

优化2(根据文末参考资料[1]中MPUCoder 的回答改写):

public class Solution
{
    public bool JudgeSquareSum(int c)
    {           
        for (int i = 0; i <= c && c - i * i >= 0; i++)
        {
            int diff = c - i*i;
            if (IsPerfectSquare(diff))
                return true;
        }

        return false;
    }
    public bool IsPerfectSquare(int num)
    {
        if ((0x0213 & (1 << (num & 15))) != 0)  //TRUE only if n mod 16 is 0, 1, 4, or 9
        {
            int t = (int)Math.Floor(Math.Sqrt((double)num) + 0.5);
            return t * t == num;
        }
        return false;
    }
}

Rank:

执行用时: 44 ms, 在所有 csharp 提交中击败了100.00%的用户.

优化3(根据文末参考资料[1]中 Simon 的回答改写):

public class Solution
{
    public bool JudgeSquareSum(int c)
    {           
        for (int i = 0; c - i * i >= 0; i++)
        {
            long diff = c - i*i;
            if (IsSquareFast(diff))
                return true;
        }

        return false;
    }

    bool IsSquareFast(long n)
    {
        if ((0x2030213 & (1 << (int)(n & 31))) > 0)
        {
            long t = (long)Math.Round(Math.Sqrt((double)n));
            bool result = t * t == n;
            return result;
        }
        return false;
    }
}

Rank:

执行用时: 48 ms, 在所有 csharp 提交中击败了100.00%的用户.

另外，stackoverflow上还推荐了一种写法：

public class Solution
{
    public bool JudgeSquareSum(int c)
    {           
        for (int i = 0; c - 2 * i * i >= 0; i++)
        {
            double diff = c - i*i;
            if (Math.Abs(Math.Sqrt(diff) % 1) < 0.000001)
                return true;
        }

        return false;
    }
}

事实上，速度并不快~

Rank:

执行用时: 68 ms, 在所有 csharp 提交中击败了27.27%的用户.

相应代码已经上传到github:

https://github.com/yanglr/Leetcode-CSharp/tree/master/leetcode633

参考资料:

[1] Fast way to test whether a number is a square
https://www.johndcook.com/blog/2008/11/17/fast-way-to-test-whether-a-number-is-a-square/

[2] Shortest way to check perfect Square? - C#
https://stackoverflow.com/questions/4885925/shortest-way-to-check-perfect-square/4886006#4886006