[LeetCode] Redundant Connection 冗余的连接,LeetCode All in One 题目讲解汇总(持续更新中...)

 

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 directedgraph 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,让我们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。我们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,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,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,所以没法采用利用变量pre来避免上面所说的死循环(不是很确定,可能是博主没想出来,有做出来的请在评论区贴上代码),所以我们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,否则我们就把该结点加入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,首先我们建立一个长度为(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,我们加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,我们懒得转换了,就多加一位得了。我们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,如果我们此时新加一条边[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,那么我们要将结点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;
    }
};

 

类似题目:

Graph Valid Tree

 

参考资料:

https://discuss.leetcode.com/topic/104729/10-line-java-solution-union-find

 

LeetCode All in One 题目讲解汇总(持续更新中…)

    原文作者:Grandyang
    原文地址: https://www.cnblogs.com/grandyang/p/7628977.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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