LeetCode-207. Course Schedule [C++][Java]-优快云博客

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LeetCode-207. Course ScheduleLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/course-schedule/

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= ai, bi < numCourses
All the pairs prerequisites[i] are unique.

【C++】

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses,  vector<int>());
        vector<int> indegree(numCourses, 0);
        for (const auto& prerequisite : prerequisites) {
            graph[prerequisite[1]].push_back(prerequisite[0]);
            ++indegree[prerequisite[0]];
        }
        queue<int> q;
        for (int i = 0; i < indegree.size(); i++) {
            if (!indegree[i]) {q.push(i);}
        }
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (auto v : graph[u]) {
                --indegree[v];
                if (!indegree[v]) {q.push(v);}
            }
        }
        for (int i = 0; i < indegree.size(); i++) {
            if (indegree[i]) {return false;}
        }
        return true;
    }
};

【Java】

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[] inDegree = new int[numCourses];
        List<List<Integer>> graph = new ArrayList<>(numCourses);
        for (int i = 0; i < numCourses; ++i) {graph.add(new LinkedList<>());}
        for (int[] prerequisite: prerequisites) {
            graph.get(prerequisite[1]).add(prerequisite[0]);
            ++inDegree[prerequisite[0]];
        }

        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < inDegree.length; ++i) {
            if (inDegree[i] == 0) {q.add(i);}
        }
        int size = 0;
        while (!q.isEmpty()) {
            int u = q.poll();
            size++;
            for (int v: graph.get(u)) {
                --inDegree[v];
                if (inDegree[v] == 0) {q.add(v);}
            }
        }
        return size == numCourses;
    }
}