题目：原题链接（困难）
标签：并查集、贪心算法、图
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 40ms (75.86%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class DSU1:
def __init__(self, n: int):
self._n = n
self._array = [i for i in range(n)]
self._size = [1] * n
self.group_num = n # 连通分支数量
def find(self, i: int) -> int:
if self._array[i] != i:
self._array[i] = self.find(self._array[i])
return self._array[i]
def union(self, i: int, j: int) -> bool:
i, j = self.find(i), self.find(j)
if i != j:
if self._size[i] >= self._size[j]:
self._array[j] = i
self._size[i] += self._size[j]
else:
self._array[i] = j
self._size[j] += self._size[i]
self.group_num -= 1
return True
else:
return False
def is_connected(self, i: int, j: int) -> bool:
return self.find(i) == self.find(j)
def get_size(self, i):
return self._size[self.find(i)]
def __repr__(self):
return str(len(self._array)) + ":" + str(self._array)
class Solution:
def minSwapsCouples(self, row: List[int]) -> int:
size = len(row)
dsu = DSU1(size // 2)
for i in range(0, size, 2):
dsu.union(row[i] // 2, row[i + 1] // 2)
count = set()
for i in range(size // 2):
count.add(dsu.find(i))
return size // 2 - len(count)
