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;
}
}
};