题目：原题链接（中等）
标签：并查集、图
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( N ) | 80ms (6.89%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class DSU2:
def __init__(self):
self._n = 0
self._parent = {}
self._size = {}
def __contains__(self, i):
return i in self._parent
def add(self, i):
if i not in self._parent:
self._parent[i] = i
self._size[i] = 1
def get_size(self, i):
return self._size[self.find(i)]
def find(self, i):
if self._parent[i] != i:
self._parent[i] = self.find(self._parent[i])
return self._parent[i]
def union(self, i, j):
i, j = self.find(i), self.find(j)
if i != j:
self._parent[i] = j
self._size[j] += self._size[i]
del self._size[i]
def is_connected(self, x, y):
return self.find(x) == self.find(y)
class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
dsu = DSU2()
for equation in equations:
if "==" in equation:
n1, n2 = equation.split("==")
if n1 not in dsu:
dsu.add(n1)
if n2 not in dsu:
dsu.add(n2)
dsu.union(n1, n2)
for equation in equations:
if "!=" in equation:
n1, n2 = equation.split("!=")
if n1 not in dsu:
dsu.add(n1)
if n2 not in dsu:
dsu.add(n2)
if dsu.find(n1) == dsu.find(n2):
return False
return True
