Medium
首先还是一样，阅读理解 先看等式，a==b , b==c 从此，我们可以得到隐藏的条件a == c，这时假如数组中有a != c，那就是矛盾的。
==符号链接的元素，形成一个树状结构!=符号连接，那就返回false综上，AC！
python3
class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
class UnionFind():
def __init__(self, equations):
self.s = {}
for e in equations:
self.s[e[0]] = e[0]
self.s[e[3]] = e[3]
def find(self, x):
while x != self.s[x]:
x = self.s[x]
return x
def union(self, a ,b):
if self.find(a) != self.find(b):
self.s[self.find(a)] = self.find(b)
uf = UnionFind(equations)
for e in equations:
if e[1] != "!":
uf.union(e[0], e[3])
for e in equations:
if e[1] == "!":
if uf.find(e[0]) == uf.find(e[3]):
return False
return True