684. Redundant connection
In this problem, the tree refers to an undirected graph that is connected and acyclic .
Enter a graph, which consists of a tree with N nodes (the node values do not repeat 1, 2, ..., N) and an additional edge. The two vertices of the additional edge are contained between 1 and N. This additional edge does not belong to an existing edge in the tree.
The resulting graph is a 边two-dimensional array. Each 边element is a pair [u, v] , satisfying u < v, which represents the edge of the undirected graph connecting the vertices u and .v
Return an edge that can be deleted, making the resulting graph a tree with N nodes. If there are multiple answers, the last edge in the two-dimensional array is returned. The answer side [u, v]should meet the same format u < v.
Example 1:
Input: [[1,2], [1,3], [2,3]] Output: [2,3] Explanation: The given undirected graph is: 1 / \ 2-3
Example 2:
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph is:
5-1-2
| |
4-3
note:
- The size of the input 2D array is from 3 to 1000.
- The integers in the two-dimensional array are between 1 and N, where N is the size of the input array.
class Solution(object):
def findRedundantConnection(self, edges):
"""
:type edges: List[List[int]]
:rtype: List[int]
"""
self.init(len(edges))
ans = []
for u,v in edges:
is_success, e2 = self.connect(u, v)
if not is_success:
ans = e2
return ans
def init(self, N):
self.father = {}
for i in range(1, N+1):
self.father[i] = i
def connect(self, u, v):
f1 = self.find_father(u)
f2 = self.find_father(v)
if f1 == f2:
return False, [u, v]
else:
self.father[f1] = f2
return True, []
def find_father(self, u):
root = u
path = []
while self.father[root] != root:
path.append(root)
root = self.father[root]
for p in path:
self.father[p] = root
return root