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
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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
原文地址:https://www.cnblogs.com/enjoy233/p/csharp_leetcode_series_4.html