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

淡淡的烟草味﹌ 2022-06-03 08:28 143阅读 0赞

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;
            }
        }
    };