【leetcode刷题】T208-平方数之和

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

i从0到sqrt(num)+1，判断i ** 2 + sqrt(num-i*i) ** 2是否等于num
另一个想法，生成0到sqrt(num)+1所有数的平方，那么问题就变成了two-sum：在数组中，找到两个元素，使之和为num

class Solution(object):
    def judgeSquareSum(self, c):
        """
        :type c: int
        :rtype: bool
        """
        for i in range(int(math.sqrt(c)) + 1):
            tmp1 = i * i
            tmp2 = int(round(math.sqrt(c - tmp1))) ** 2
            if tmp1 + tmp2 == c:
                return True
        return False

class Solution(object):
    def judgeSquareSum(self, c):
        ls = [i ** 2 for i in range(int(math.sqrt(c)) + 1)]
        a, b = 0, len(ls) - 1
        while a <= b:
            if ls[a] + ls[b] == c:
                return True
            if ls[a] + ls[b] > c:
                b -= 1
            else:
                a += 1
        return False