685.冗余连接II

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题目链接：685.冗余连接II

代码如下：

//并查集
//参考链接：https://leetcode.cn/problems/redundant-connection-ii/solutions/417164/685-rong-yu-lian-jie-iibing-cha-ji-de-ying-yong-xi
class Solution 
{
public:vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {int inDegree[N]={0};n=edges.size();for(int i=0;i<edges.size();i++){inDegree[edges[i][1]]++;//统计入度}vector<int> vec;//记录入度为2的边for(int i=edges.size()-1;i>=0;i--){if(inDegree[edges[i][1]]==2){vec.push_back(i);}}if(vec.size()>0){if(isTreeAfterRemoveEdge(edges,vec[0])){return edges[vec[0]];}else{return edges[vec[1]];}}return getRemoveEgdge(edges);}
private:static const int N = 1010;vector<int> father=vector<int>(N, 0);int n;//边的数量//并查集初始化void init(){for(int i=1;i<=n;i++){father[i]=i;}}//并查集寻根int find(int u){return u==father[u]?u:father[u]=find(father[u]);//路径压缩}//并查集判定u，v是否是一个根bool isSame(int u,int v){u=find(u);v=find(v);return u==v;}//将v->v这条边加入并查集void join(int u,int v){u=find(u);v=find(v);if(u==v)    return;father[v]=u;}//在有向图中找到删除的那条边，使其成为一棵树vector<int> getRemoveEgdge(const vector<vector<int>>& edges){   init();for(int i=0;i<edges.size();i++){if(isSame(edges[i][0],edges[i][1]))//构成了有向环，就需要删除{return edges[i];}join(edges[i][0],edges[i][1]);}return {};}//删除一条边后判断是否是树bool isTreeAfterRemoveEdge(const vector<vector<int>>& edges,int deleteEdge){init();for(int i=0;i<edges.size();i++){if(i==deleteEdge)   continue;if(isSame(edges[i][0],edges[i][1]))//构成了有向环，一定不是树{return false;}join(edges[i][0],edges[i][1]);}return true;}
};