leetcode 633. Sum of Square Numbers 二分查找+勾股定理

Given a non-negative integer c, your task is to decide whether there’re two integers a and b such that a2 + b2 = c.

Example 1:

Input: 5

Output: True

Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3

Output: False

本题很简单，就是寻找勾股数，我最初的想法是直接暴力求解但是发现太笨了，后来在网上看到了一个二分查找的做法，十分棒，值得学习

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

class Solution
{
public:
bool judgeSquareSum(int c)
{
if (c < 0)
return false;
int left = 0, right = (int)sqrt(c);
while (left <= right)
{
int res = left*left + right*right;
if (res < c)
left++;
else if (res > c)
right--;
else
return true;
}
return false;
}
};