leetcode 684. Redundant Connection 邻接表的环的判断 + 深度优先遍历DFS

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.

本题题意很简单，就是寻找组成环的第一条变，那么做法就是每加入一条边，就进行环检测，一旦发现了环，就返回当前边。对于无向图，我们还是用邻接表来保存，建立每个结点和其所有邻接点的映射，由于两个结点之间不算有环，所以我们要避免这种情况 1->{2}, 2->{1}的死循环，所以我们用一个变量pre记录上一次递归的结点，比如上一次遍历的是结点1，那么在遍历结点2的邻接表时，就不会再次进入结点1了，这样有效的避免了死循环，使其能返回正确的结果，

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution {public:    map<int, set<int>> graph;    vector<int> findRedundantConnection(vector<vector<int>>& edges)    {        for (auto edge : edges)        {            if (hasCycle(edge[0], edge[1], -1) == true)                return edge;            else            {                graph[edge[0]].insert(edge[1]);                graph[edge[1]].insert(edge[0]);            }        }        return{};    }    bool hasCycle(int cur, int target, int pre)    {        if (graph[cur].find(target) != graph[cur].end())            return true;        else        {            for (int a : graph[cur])            {                if (a == pre)                    continue;                if (hasCycle(a, target, cur) == true)                    return true;            }            return false;        }    }};