Topological Sorting

Description

Given an directed graph, a topological order of the graph nodes is defined as follow:

For each directed edge A -> B in graph, A must before B in the order list. The first node in the order can be any node in the graph with no nodes direct to it. Find any topological order for the given graph.

Notice
You can assume that there is at least one
topological order in the graph.

Clarification
Learn more about representation of graphs

Example
For graph as follow:
The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

/**
 * Definition for Directed graph.
 * struct DirectedGraphNode {
 *     int label;
 *     vector<DirectedGraphNode *> neighbors;
 *     DirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    /**
     * @param graph: A list of Directed graph node
     * @return: Any topological order for the given graph.
     */
    vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {
        // write your code here
        if (graph.empty()) {
            return {};
        }
        vector<DirectedGraphNode*> result;
        int len = graph.size();
        unordered_map<DirectedGraphNode*, int> indegree;
        // find all nodes with 0 indegree
        for (int i = 0; i < len; ++i) {
            // search all neighbors
            for (auto& neighbor : graph[i]->neighbors) {
                // cannot find
                if (indegree.find(neighbor) == indegree.end()) {
                    indegree[neighbor] = 1;
                } else {
                // can find
                    indegree[neighbor]++;
                }
            }
        }
        // init q with indegree 0, namely, indegree cannot find
        queue<DirectedGraphNode*> q;
        for (const auto& entry : graph) {
            if (indegree.find(entry) != indegree.end()) {
                continue;
            }
            q.push(entry);
        }
        // bfs
        while (!q.empty()) {
            // DirectedGraphNode* cur = q.front();
            auto& cur = q.front();
            q.pop();
            result.push_back(cur);
            for (const auto& i : cur->neighbors) {
                indegree[i]--;
                if (indegree[i] == 0) {
                    q.push(i);
                }
            }
        }
        return result;
    }
};

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