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 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

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;
    }
};