C#版 - Leetcode 633. 平方数之和 - 题解
C#版 - Leetcode 633. 平方数之和 - 题解
C#版 - Leetcode 633. 平方数之和 - 题解
Leetcode 633 - Sum of square number
在线提交: 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 a2+b2=ca2+b2=ca^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: FalseInput: 5 2 100
Expected answer: true true true
- 题目难度： 简单
- 通过次数：1.1K
- 提交次数：4.8K
- 贡献者： Stomach_ache
- 相关话题 数学 相似题目 有效的完全平方数
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。循环变量i满足 i2⋅2<c2i2⋅2<c2i^2 \cdot 2 < c^2 .
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))))
return true;
}
return false;
}
}Rank:
You are here! Your runtime beats 56.14% of csharp submissions. 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: You are here! Your runtime beats 85.96% of csharp submissions.
优化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.
优化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: 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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