這次是檢查冗餘的,題目主要練習無向圖中是否有環的判斷。
題目如下:
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.
代碼如下:
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
unordered_map<int, unordered_set<int>> m;
for (auto x : edges) {
if (ifCycle(x[0], x[1], m, -1)) return x;
m[x[0]].insert(x[1]);
m[x[1]].insert(x[0]);
}
return {};
}
bool ifCycle(int cur, int des, unordered_map<int, unordered_set<int>>&m, int pre) {
if (m[cur].count(des)) return true;
for (int num : m[cur]) {
if (num == pre) continue;
if (ifCycle(num, des, m, cur)) return true;
}
return false;
}
};