Redundant Connection——LeetCode进阶路

原题链接https://leetcode.com/problems/redundant-connection/

题目描述

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.

思路分析

题目要求很简单，就是要把给出的无向图的最后一条边删掉
和之前的解法类似，对遍历过的元素标记，发现环时,删掉最后元素

AC解

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        if(edges == null || edges.length == 0 || edges[0].length == 0)
        {
            return new int[2];
        }
        
        ArrayList<Integer>[] list = new ArrayList[edges.length + 1];
        for(int i=0; i<list.length; i++)
        {
            list[i] = new ArrayList<Integer>();
        }
        
        for(int i=0; i<edges.length; i++)        
        {
            int u = edges[i][0];
            int v = edges[i][1];
            list[u].add(v);
            list[v].add(u);
            
            boolean[] flag = new boolean[edges.length + 1];
            if(!dfs(u,list,-1,flag))
            {
                return edges[i];
            }
        }
        return new int[2];
    }
    
    public boolean dfs(int u, ArrayList<Integer> list[],int last, boolean[] flag)
    {
        if(flag[u])
        {
            return false;
        }
        
        flag[u] = true;
        for(int i:list[u])
        {
            if(i != last)
            {
                if(!dfs(i,list,u,flag))
                {
                    return false;
                }
            }
        }
        return true;
    }
}