[leetcode] 684. Redundant Connection

Description

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.

分析

题目的意思是：给定一个无向图，删掉组成环的最后一条边。

• 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v[1] = 2，就表示结点1和结点2是相连的，v[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v[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v[1] = 3即可

代码

class Solution {
public:
    vector findRedundantConnection(vector>& edges) {
        vector v(2001,-1);
        for(auto edge:edges){
            int x=find(v,edge[0]);
            int y=find(v,edge[1]);
            if(x==y) return edge;
            v[x]=y;
        }
        return {};
    }
    int find(vector v,int i){
        while(v[i]!=-1){
            i=v[i];
        }
        return i;
    }
};

参考文献

[LeetCode] Redundant Connection 冗余的连接