leetcode 367. Valid Perfect Square

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

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

经典的二分查找：

class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        # find n in (1, num) that n*n == num
        # binary search
        i, j = 1, num
        while i <= j:
            mid = (i+j)>>1
            s = mid*mid
            if s == num:
                return True
            elif s > num:
                j = mid-1
            else:
                i = mid+1
        return False

 其他解法：

class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        x = int(round(math.exp(math.log(num)/2)))
        return x*x == num

 注意，int（round（xxx））是经典写法，记住！

另外的解法：A square number is 1+3+5+7+..., JAVA code，不知道这个是否数学证明过。。。

class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        i = 1
        while num > 0:
            num -= i
            i += 2
        return num == 0

 最后就是牛顿法求解了。

class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        x = num;
        while (x * x > num):
            x = (x + num / x) >> 1        
        return x * x == num

但是不知道是否严密，因为num/x是整数相除，我自己倾向于下面这样解：

class Solution(object):
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        prev, cur = 0, num
        while abs(prev-cur)>1e-6:
            cur,prev=0.5*(cur+num/cur),cur
        x = int(round(cur))
        return x*x == num