## Description

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

## Solutions

判断一个数是不是两个数的平方和。

### 1. Brute Force

第一想法写出来的代码时间复杂度是 $O(n^2)$，过不了时间限制：

class Solution(object):
def judgeSquareSum(self, c):
"""
:type c: int
:rtype: bool
"""
from math import sqrt

if c == 0 or c == 1:
return True

n = int(sqrt(c))
for i in range(n+1):
res = c - i ** 2
n_res = int(sqrt(res))
for j in range(n_res+1):
if j ** 2 == res:
return True
return False

### 2. Two Pointers

参考了别人做法，确实快好多。

class Solution(object):
def judgeSquareSum(self, c):
"""
:type c: int
:rtype: bool
"""
from math import sqrt
left, right = 0, int(sqrt(c))
while left <= right:
squre_sum = left * left + right * right
if squre_sum < c:
left += 1
elif squre_sum > c:
right -= 1
else:
return True
return False
# Runtime: 76 ms, faster than 95.89% of Python online submissions for Sum of Square Numbers.
# Memory Usage: 11.7 MB, less than 83.97% of Python online submissions for Sum of Square Numbers.