一个无向图是否存在环
算法1：
我们知道对于环 1-2-3-4-1，每个节点的度都是2，基于此我们有如下算法（这是类似于有向图的拓扑排序）：
- 求出图中所有顶点的度
- 删除图中所有度 <=1 的顶点以及与该顶点相关的边，把与这些边相关的顶点的度减一
- 如果还有度<=1的顶点重复步骤2
- 最后如果还存在未被删除的顶点，则表示有环；否则没有环
时间复杂度为O(E+V)，其中E、V分别为图中边和顶点的数目。
算法2：
深度优先遍历该图，如果在遍历的过程中，发现某个节点有一条边指向已经访问过的节点，并且这个已访问过的节点不是当前节点的父节点（这里的父节点表示dfs遍历顺序中的父节点），则表示存在环。但是我们不能仅仅使用一个bool数组来标志节点是否访问过。
对每个节点分为三种状态，白、灰、黑。
开始时所有节点都是白色，当开始访问某个节点时该节点变为灰色，当该节点的所有邻接点都访问完，该节点颜色变为黑色。
那么我们的算法则为：如果遍历的过程中发现某个节点有一条边指向颜色为灰的节点，那么存在环。
LeeCode题目
LeetCode题目：684. Redundant Connection
In this problem, a tree is an undirected 无向图 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:
Example 1
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:
Example 2
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.
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
//int eSize = edges.size();
vector<int> start(3000);
//memset(start, 0, sizeof(int)*eSize);
for (int i = 0; i < 3000; i++) {
start[i] = i;
}
int tStart, tEnd;
for (int i = 0; i < edges.size(); i++) {
tStart = edges[i][0];
tEnd = edges[i][1];
if (start[tStart] == start[tEnd]) {
return edges[i];
}
int needUpdate = start[tEnd];
for (int j = 0; j < 3000; j++) {
if (start[j] == needUpdate) {
start[j] = start[tStart];
}
}
}
}
};
Note：上述算法的时间复杂度是O(n^2)