[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

• 開始表示每個結點都是一個單獨的組，所謂的Union Find就是要讓結點之間建立關聯，比如若v[1] = 2，就表示結點1和結點2是相連的，v[2] = 3表示結點2和結點3是相連的，如果我們此時新加一條邊[1, 3]的話，我們通過root[1]得到2，再通過v[2]得到3，說明結點1有另一條路徑能到結點3，這樣就說明環是存在的；如果沒有這條路徑，那麼我們要将結點1和結點3關聯起來，讓v[1] = 3即可

代碼

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        vector<int> 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<int> v,int i){
        while(v[i]!=-1){
            i=v[i];
        }
        return i;
    }
};      

參考文獻