210. Course Schedule II (Medium)

https://leetcode.com/problems/course-schedule-ii/

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: 2, [[1,0]] 
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished   
             course 0. So the correct course order is [0,1] .

Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both     
             courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. 
             So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .

Note:

  1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
  2. You may assume that there are no duplicate edges in the input prerequisites.

Hints

Solutions

class Solution {
  static int WHITE = 1;
  static int GRAY = 2;
  static int BLACK = 3;

  boolean isPossible;
  Map<Integer, Integer> color;
  Map<Integer, List<Integer>> adjList;
  List<Integer> topologicalOrder;

  private void init(int numCourses) {
    this.isPossible = true;
    this.color = new HashMap<Integer, Integer>();
    this.adjList = new HashMap<Integer, List<Integer>>();
    this.topologicalOrder = new ArrayList<Integer>();

    // By default all vertces are WHITE
    for (int i = 0; i < numCourses; i++) {
      this.color.put(i, WHITE);
    }
  }

  private void dfs(int node) {

    // Don't recurse further if we found a cycle already
    if (!this.isPossible) {
      return;
    }

    // Start the recursion
    this.color.put(node, GRAY);

    // Traverse on neighboring vertices
    for (Integer neighbor : this.adjList.getOrDefault(node, new ArrayList<Integer>())) {
      if (this.color.get(neighbor) == WHITE) {
        this.dfs(neighbor);
      } else if (this.color.get(neighbor) == GRAY) {
        // An edge to a GRAY vertex represents a cycle
        this.isPossible = false;
      }
    }

    // Recursion ends. We mark it as black
    this.color.put(node, BLACK);
    this.topologicalOrder.add(node);
  }

  public int[] findOrder(int numCourses, int[][] prerequisites) {

    this.init(numCourses);

    // Create the adjacency list representation of the graph
    for (int i = 0; i < prerequisites.length; i++) {
      int dest = prerequisites[i][0];
      int src = prerequisites[i][1];
      List<Integer> lst = adjList.getOrDefault(src, new ArrayList<Integer>());
      lst.add(dest);
      adjList.put(src, lst);
    }

    // If the node is unprocessed, then call dfs on it.
    for (int i = 0; i < numCourses; i++) {
      if (this.color.get(i) == WHITE) {
        this.dfs(i);
      }
    }

    int[] order;
    if (this.isPossible) {
      order = new int[numCourses];
      for (int i = 0; i < numCourses; i++) {
        order[i] = this.topologicalOrder.get(numCourses - i - 1);
      }
    } else {
      order = new int[0];
    }

    return order;
  }
}
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