684. Redundant Connection

QUESTION:

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.

EXPLANATION:

这还是一个union-find的题目，union-find的算法一书中就说过，如果两个点本身就连在一起，那么我们就不将他们连在一起。而这个题目也就是将原本就连在一起的进行拆分。
思路：首先采用union的方式，将连接的点都连接在一起，如1，2两点，那么就都改成2，2，直到找到一个本身就连接在一起的点，也就是本来是（n,n）的点，就说明这个是需要remove的地方。
逻辑：
1.找到数组中最大的数max
2.创建一个长度为max+1数组，并且初始化每个位置为当前位置值
3.对edges进行连线操作，并且设置对应的parent
4.如果找到当前的两点的parent就是相同的，说明两者本来就连在一起，就是我们需要找的值
5.一直找到最后一对，返回即可

SOLUTION:

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int max = 0;
        for(int i = 0;i<edges.length;i++){
            for(int x: edges[i]){
                if(x>max)max = x;
            }
        }
        int[] index = new int[max+1];
        int[] result = new int[2];
        for(int i = 0;i<index.length;i++) index[i] = i;
        for(int i = 0;i<edges.length;i++){
            if(findRedundantConnectionHelper(index,edges[i])) result = edges[i];
        }
        return result;
    }
    public static boolean findRedundantConnectionHelper(int[] index,int[] pair){
        int a = pair[0];
        int b = pair[1];
        int ida = index[a];
        int idb = index[b];
        if(index[a] == index[b]) return true;
        for(int i = 0;i<index.length;i++){
            if(index[i]==ida) index[i] = idb;
        }
        return false;
    }
}
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