[LintCode] Find the Weak Connected Component in the Directed Graph

Find the number Weak Connected Component in the directed graph. Each node in the graph contains a label and a list of its neighbors. (a connected set of a directed graph is a subgraph in which any two vertices are connected by direct edge path.)

Example

Given graph:

A----->B  C
\     |  |
\    |  |
\   |  |
\  v  v
->D  E <- F

Return

{A,B,D}, {C,E,F}

. Since there are two connected component which are

{A,B,D} and {C,E,F}

Note

Sort the element in the set in increasing order.

Solution:

并查集。遍历每一条变，更新每个节点所在集合。

/**
* Definition for Directed graph.
* struct DirectedGraphNode {
*     int label;
*     vector<DirectedGraphNode *> neighbors;
*     DirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
//use union-set to solve
private:
int find(unordered_map<int, int> &nodeMap, int label){
if(nodeMap.find(label) == nodeMap.end()){
nodeMap[label] = label;
return label;
//if this node doesn't belong to any union-set, create a new set
}else{
//this node belongs to some set, find the root of the set
int res = nodeMap[label];
while(nodeMap[res] != res)
res = nodeMap[res];
return res;
}
}
public:
/**
* @param nodes a array of directed graph node
* @return a connected set of a directed graph
*/
vector<vector<int>> connectedSet2(vector<DirectedGraphNode*>& nodes) {
unordered_map<int, int> nodeMap;
vector<vector<int> > result;
for(int i = 0;i < nodes.size();++i){
for(int j = 0;j < (nodes[i]->neighbors).size();++j){
int s1 = find(nodeMap, nodes[i]->label);
int s2 = find(nodeMap, (nodes[i]->neighbors)[j]->label);
if(s1 != s2){
//union two sets
if(s1 < s2) nodeMap[s2] = s1;
else nodeMap[s1] = s2;
}else{
//do nothing
}
}
}

unordered_map<int, int> setId2VecId;
for(int i = 0;i < nodes.size();++i){
int label = nodes[i]->label;
int setId = find(nodeMap, label);
if(setId2VecId.find(setId) == setId2VecId.end()){
vector<int> vec;
setId2VecId[setId] = result.size();
result.push_back(vec);
}
int idx = setId2VecId[setId];
result[idx].push_back(label);
}
for(int i = 0;i < result.size();++i)
sort(result[i].begin(), result[i].end());

return result;
}
};

Inspired by: http://www.cnblogs.com/easonliu/p/4607300.html