C# LeetCode刷題 - Leetcode 633. 平方數之和 - 題解

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C#版 - Leetcode 633. 平方數之和 - 題解

Leetcode 633 - Sum of square number

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

題目描述

給定一個非負整數 c ，你要判斷是否存在兩個整數 a和 b，使得 $a^2 + b^2 = c$。

示例1:

輸入: 5
輸出: True
解釋: 1 * 1 + 2 * 2 = 5

示例2:

輸入: 3
輸出: False

Input:
5
2
100

Expected answer:
true
true
true

---

●  題目難度:

簡單

- 通過次數：1.1K - 提交次數：4.8K - 貢獻者： [Stomach_ache](https://leetcode.com/stomach_ache/)

• 相關話題 數學
  相似題目 有效的完全平方數

---

思路：

做一次循環，用目標和減去循環變量的平方，如果剩下的部分依然是完全平方的情形存在，就返回true，否則返回false。循環變量i滿足 $i^2 \cdot 2 < c^2 $.

已AC代碼:
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))))
                return true;
        }

        return false;
    }
}

Rank:

You are here! Your runtime beats 56.14% of csharp submissions.


優化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:
You are here! Your runtime beats 85.96% of csharp submissions.

優化2:

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)
        {
            int t = (int)Math.Floor(Math.Sqrt((double)num) + 0.5);
            return t * t == num;
        }
        return false;
    }
}

Rank:
You are here! Your runtime beats 85.96% of csharp submissions.

優化3:

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:
You are here! Your runtime beats 85.96% of csharp submissions.

另外，stackoverflow上還推薦了一種寫法：

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;
    }
}

事實上，速度並不快：
Rank:
You are here!
Your runtime beats 29.82% of csharp submissions.

Reference:

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/

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