# 684. Redundant Connection

In this problem, a tree is an undirected graph that is connected and has no cycles.

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.

``````class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
unordered_map<int, unordered_set<int>> m;
for (auto edge : edges) {
if (hasCycle(edge[0], edge[1], m, -1)) return edge;
m[edge[0]].insert(edge[1]);
m[edge[1]].insert(edge[0]);
}
return {};
}
bool hasCycle(int cur, int target, unordered_map<int, unordered_set<int>>& m, int pre) {
if (m[cur].count(target)) return true;
for (int num : m[cur]) {
if (num == pre) continue;
if (hasCycle(num, target, m, cur)) return true;
}
return false;
}
};
``````

``````class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
unordered_map<int, unordered_set<int>> m;
for (auto edge : edges) {
queue<int> q{{edge[0]}};
unordered_set<int> s{{edge[0]}};
while (!q.empty()) {
auto t = q.front(); q.pop();
if (m[t].count(edge[1])) return edge;
for (int num : m[t]) {
if (s.count(num)) continue;
q.push(num);
s.insert(num);
}
}
m[edge[0]].insert(edge[1]);
m[edge[1]].insert(edge[0]);
}
return {};
}
};
``````

``````class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
vector<int> root(2001, -1);
for (auto edge : edges) {
int x = find(root, edge[0]), y = find(root, edge[1]);
if (x == y) return edge;
root[x] = y;
}
return {};
}
int find(vector<int>& root, int i) {
while (root[i] != -1) {
i = root[i];
}
return i;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/684

Friend Circles

Accounts Merge

Redundant Connection II

Number of Islands II

Graph Valid Tree

Number of Connected Components in an Undirected Graph

Similar String Groups

https://leetcode.com/problems/redundant-connection/

https://leetcode.com/problems/redundant-connection/discuss/112562/My-DFS-and-BSF-solutions

https://leetcode.com/problems/redundant-connection/discuss/107984/10-line-Java-solution-Union-Find.

https://leetcode.com/problems/redundant-connection/discuss/108010/C%2B%2B-solution-using-union-find

LeetCode All in One 题目讲解汇总(持续更新中…)

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