leetcode 684. Redundant Connection

写在前面

contest 51 的第三题。这题如果知道并查集，基本上写出来就AC了，关键还是对并查集的理解和熟悉程度。

题目描述

We are given a “tree” in the form of a 2D-array, with distinct values for each node.

In the given 2D-array, each element pair [u, v] represents that v is a child of u in the tree.

We can remove exactly one redundant pair in this “tree” to make the result a tree.

You need to find and output such a pair. If there are multiple answers for this question, output the one appearing last in the 2D-array. There is always at least one answer.

Example 1:
Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: Original tree will be like this:
1
/ \
2 - 3
Example 2:
Input: [[1,2], [1,3], [3,1]]
Output: [3,1]
Explanation: Original tree will be like this:
1
/ \
2 3
Note:
The size of the input 2D-array will be between 1 and 1000.
Every integer represented in the 2D-array will be between 1 and 2000.

思路分析

简单讲就是在图中找出多余的边，该边对已存在的结点连通性没有任何影响，即，边上的两个结点已经是连通状态（connected）。这其实跟并查集的定义非常相近，我们只需要不断union未连通的结点，遇到connected的结点，return这条边即可。关于并查集，请仔细阅读这篇博客（并查集介绍），最好能够自己写出并查集代码。

代码实现

AC代码中的并查集实现基本上以上述博客为准，代码中所做的操作就是简单的union和判断是否连通（connected）。

class Solution {    public:    class UF {        private:        vector<int> id;        vector<int> sz;        int c;        public:        UF(int N):c(N) {            for(int i = 0;i<N;++i) {                id.push_back(i);                sz.push_back(1);            }        }        int count(){return c;};        bool connected(int p,int q) {return find(p) == find(q);}        int find(int p) {            while(p!=id[p]) {                id[p] = id[id[p]];                p = id[p];            }            return p;        }        void UN(int p,int q) {            int i = find(p);            int j = find(q);            if(i==j) return;            if(sz[i]<sz[j]) {                id[i] = j;                sz[j] += sz[i];            }            else {                id[j] = i;                sz[i]+=sz[j];            }            c--;        }    };public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        // 并查集        // 第一步 构建并查集        UF uf(2001);        for(auto&val:edges) {            if(uf.connected(val[0]-1,val[1]-1)) return val;            uf.UN(val[0]-1,val[1]-1);        }        return {};    }};