Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Returns: True

Example 2:

Input: 14
Returns: False

思路：所謂perfect Square，A perfect square is a number that can be expressed as the** product of two equal integers**.

可以參照perfect number的思路，從1到sqrt(num)依次比對，后來發(fā)現(xiàn)會提前跳出循環(huán)。突然靈機一動，想出了個取巧的方法。使用Python的**0.5代替sqrt，能被accept，不過好像還是不滿足題目要求。

看了其他人的思路，發(fā)現(xiàn)還有一個牛頓法，以及二分法查找。二分法查找倒是非常直觀。

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        return (int(num**0.5)**2)==num

    def isPerfectSquare2(self, num):
        res = num
        while(res*res > num):
            res = (res + num/res) >> 1

        return res*res == num

    def isPerfectSquare3(self, num):
        res = num
        while res*res > num:
            res = (res + num/res)/2
        return res*res == num

    def isPerfectSquare4(self, num):
        low = 1
        high = num
        while(low < high):
            mid = (low + high)/2
            if(mid*mid>num):
                high = mid -1
            elif(mid*mid<num):
                low = mid + 1
            else:
                return True

        return False



if __name__ == '__main__':
    sol = Solution()
    print sol.isPerfectSquare(4)
    print sol.isPerfectSquare(16)
    print sol.isPerfectSquare(14)

    print sol.isPerfectSquare(8)
    print sol.isPerfectSquare(36)

    print sol.isPerfectSquare2(16)
    print sol.isPerfectSquare3(16)

    print sol.isPerfectSquare4(16)