leetcode 684. Redundant Connection (union-find算法实现)

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.

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加权的union-find算法，记录每个根节点dep，这样算法find步骤就不会从对数复杂度？退化为线性复杂度。

时间复杂度分析见 https://leetcode.com/problems/redundant-connection/solution/

union-find的思想和dfs时的染色差不多，边遍历，边标记。

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        UF uf=new UF(edges.length+1);
        for(int[] e:edges){
            if(uf.union(e[0],e[1])){
               return e; 
            }
        }
        throw null;
    }
    class UF{
        int[] id;//节点i联通id[i]，一直连接到根节点，为联通的标识
        int[] dep;//根节点深度
        UF(int N){
            id=new int[N];
            dep=new int[N];
            for(int i=0;i<N;i++) id[i]=i;
        }
        
        public int find(int n){
            while(n!=id[n]) n=id[n];
            return n;
        }
        
        public boolean union(int p,int q){
            int pRoot=find(p);
            int qRoot=find(q);
            if(pRoot==qRoot) return true;
            if(dep[pRoot]<dep[qRoot]){
                id[pRoot]=qRoot;
            }else if(dep[pRoot]>dep[qRoot]){
                id[qRoot]=pRoot;
            }else{
                id[pRoot]=qRoot;
                dep[qRoot]++;//深度+1
            }
            return false;
        }
    }
}