684. Redundant Connection

置顶

Created At : 2023-10-28 15:14

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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.

这道题给我们了一个无向图，让删掉组成环的最后一条边，其实这道题跟之前那道 Graph Valid Tree 基本没什么区别，三种解法都基本相同。博主觉得老题稍微变一下就是一道新题，而 onsite 遇到原题的概率很小，大多情况下都会稍稍变一下，所以举一反三的能力真的很重要，要完全吃透一道题也不太容易，需要多下功夫。首先来看递归的解法，这种解法的思路是，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，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，所以要避免这种情况 1->{2}, 2->{1} 的死循环，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，参见代码如下：

解法一：

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;
    }
};

既然递归能做，一般来说迭代也木有问题。但是由于 BFS 的遍历机制和 DFS 不同，这里博主没有采用利用变量 pre 来避免上面所说的死循环(一定要使用的话可以将一个 pair 对儿放到队列中去，具体可以参见热心网友玻璃公主的留言)，所以采用一个集合来记录遍历过的结点，如果该结点已经遍历过了，那么直接跳过即可，否则就把该结点加入 queue 和集合，继续循环，参见代码如下：

解法二：

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 {};
    }
};

其实这道题最好的解法使用 Union Find 来做，论坛上清一色的都是用这种解法来做的，像博主用 DFS 和 BFS 这么清新脱俗的方法还真不多:) 其实 Union Find 的核心思想并不是很难理解，首先建立一个长度为 (n+1) 的数组 root，由于这道题并没有明确的说明n是多少，只是说了输入的二位数组的长度不超过 1000，那么n绝对不会超过 2000，加1的原因是由于结点值是从1开始的，而数组是从0开始的，懒得转换了，就多加一位得了。将这个数组都初始化为 -1，有些人喜欢初始化为i，都可以。开始表示每个结点都是一个单独的组，所谓的 Union Find 就是要让结点之间建立关联，比如若 root[1] = 2，就表示结点1和结点2是相连的，root[2] = 3 表示结点2和结点3是相连的，如果此时新加一条边 [1, 3] 的话，我们通过 root[1] 得到2，再通过 root[2] 得到3，说明结点1有另一条路径能到结点3，这样就说明环是存在的；如果没有这条路径，那么要将结点1和结点3关联起来，让 root[1] = 3 即可，参见代码如下：

解法三：

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

类似题目：

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

参考资料：

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