标签：creating 牛顿法 ase rip += rap mat any math
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
leetcode 367. Valid Perfect Square
标签：creating 牛顿法 ase rip += rap mat any math
原文地址：https://www.cnblogs.com/bonelee/p/9185445.html
