[LeetCode] 684. Redundant Connection

题：https://leetcode.com/problems/redundant-connection/description/

题目

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.

题目大意

一个树，多一条边，找出 使得 树产生 环的那条边。这边为 graph[][] 最后给出的。

思路

并查集。

若添加的边已经在一个集合中，说明添加这条边使树成为了环。

返回该边。

Note: 并查集可以 判断 无向图是否存在环。

class Solution {
    public int findOp(int[] unionFind_s,int x){
        int r = x;
        while(unionFind_s[r]!=r)
            r = unionFind_s[r];
        int i = x;
        while(unionFind_s[i]!=r){
            int t = unionFind_s[i];
            unionFind_s[i] = r;
            i = t;
        }
        return r;
    }
    
    public void unionOp(int[] unionFind_s,int a,int b){
        int ia = findOp(unionFind_s,a);
        int ib = findOp(unionFind_s,b);
        if(ia == ib)    return;
        unionFind_s[ib] = ia;
    }
    
    public boolean isContectedOp(int[] unionFind_s,int a,int b){
        return findOp(unionFind_s,a) == findOp(unionFind_s,b);
    }
        
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        int[]unionFind_s = new int[n+1];
        for(int i =  0;i < n ;i++)
            unionFind_s[i] = i;
        for(int []edge : edges){
            if(isContectedOp(unionFind_s,edge[0],edge[1]))
                return edge;
            unionOp(unionFind_s,edge[0],edge[1]);
        }
        return new int[]{-1,-1};
    }
}