684. Redundant Connection

Medium

In this problem, a tree is an undirected graph that is connected and has noycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
  1
 / \
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
    |   |
    4 - 3

Note:

The size of the input 2D-array will be between 3 and 1000.

Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirected\ graph. For the directed\ graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

如何去掉多余的一条边，把图变成树，那就是去掉环里的一条边。然后由于有多种答案，我们需要去掉list里最靠后的边，所以我们不如在遍历的时候就开始记录边的关系，当发现加入这条边后能构成一个环，就return这条边。那么如何确定能构成环呢？比如[1,2], [2,3], [1,3]，我们发现[1,3]这条边的两个端点都在同一个线段集([1,2], [2,3])里，那我们不如就使用查并集，这样当两条边能够连接在一起时（有一个端点一样）那我们就把他们归为一组。每次加入新边时，就查找端点u，v是否在同一组，不在说明他们构不成环，在说明他们能构成环。

查并集

1. Union-Found

class UnionFound:
    def __init__(self, totalNode):
        self.parents = []
        for i in range(totalNode):
            self.parents.append(i)
    def find(self, node:int):
        while(self.parents[node] != node):
            self.parents[node] = self.parents[self.parents[node]]
            node = self.parents[node]
        return node
    def union(self, node1, node2):
        par1 = self.find(node1)
        par2 = self.find(node2)
        if par1 != par2:
            self.parents[par1] = par2 # not parents[node1] = par2
            
    def isConnected(self, node1, node2):
        return self.find(node1) == self.find(node2)

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        if not edges: return []
        uf = UnionFound(10000) # can be improved
        for e in edges:
            if uf.isConnected(e[0], e[1]):
                return e
            else:
                uf.union(e[0], e[1])

改良版：

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        if not edges: return []
        parent = {}
        def find(v):
            while parent.get(v, v) != v:
                v = parent[v]
            return v
        
        for e in edges:
            if find(e[0]) == find(e[1]):
                return e
            else:
                parent[find(e[0])] = find(e[1])