Leetcode --- 684. Redundant Connection 判断是否有环 /并查集

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原文链接：https://www.cnblogs.com/grandyang/p/7628977.html

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本文介绍了解决图中寻找冗余连接的问题，通过两种方法：深度优先搜索和并查集，来找出导致图出现环的额外边。适用于含有N个节点的无向图，探讨了如何在加入新的边后保持图的树形结构。

684. Redundant Connection

Medium

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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.

方法一：

class Solution {
public:
	vector<int> findRedundantConnection(vector<vector<int>>& edges) {
		unordered_map<int, unordered_set<int>> m;//用邻接表来记录图
		for (auto edge : edges) {
			if (hasCircle(edge[0],edge[1],m,-1)) {//设置pre,防止 1-->2,2-->1这种情况
				
				return edge;
			}
			m[edge[0]].insert(edge[1]);
			m[edge[1]].insert(edge[0]);
		}
		return {};
	}
	bool hasCircle(int cur,int target, unordered_map<int, unordered_set<int>> m,int pre) {
		if (cur==target)return true;//说明另外一条路径也能到target!
		for (int e : m[cur]) {
			if (e == pre) continue;
			if (hasCircle(e, target, m, cur))
				return true;
		}
		return false;

	}
};

方法而：使用并查集

加入边之前判断两个端点是否属于同一组，如果属于同一组，说明两个端点之间已经有一条路径了，如果再加入这条边的话就会有环！

class Solution {//使用并查集
public:
	vector<int>v;
	vector<int> findRedundantConnection(vector<vector<int>>& edges) {
		v.assign(1002, -1);
		for (auto edge : edges) {
			int x = find(edge[0]);
			int y = find(edge[1]);
			if (x == y)return edge;
			v[y] = x;
		}
		return {};
	}
	
	int find(int i) {//返回并查集的根
		while (v[i] != -1) {
			i = v[i];
		}
		return i;
	}
	
};