LeetCode Graph：M207_Course_Schedule

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

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.

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.

click to show more hints.

package com.graph;import java.util.LinkedList;import java.util.Queue;import java.util.Stack;public class M207_Course_Schedule {int n;int[][] pre;int[] visited;int m;//方法一：根据图的DFS策略，如果遍历过程中发现环，则不能修完所有课//递归//TC=O(m^n)    public boolean canFinish_rec(int numCourses, int[][] prerequisites) {    n = numCourses;    pre = prerequisites;        if(n==0 || n==1) return true;    if(pre==null || pre.length<1) return true;        visited = new int[n];    m = pre.length;        boolean flag = true;    for(int i=0; i<n; i++) {    if(visited[i]==0) {    visited[i] = 1;    flag = DFS(i);    visited[i] = 2;    if(!flag) return false;    }    }        return flag;    }    boolean DFS(int v) {    boolean flag = true;        for(int i=0; i<m; i++) {//遍历边的时候，可以优化，将已经访问过的边标记    int a = pre[i][0];    int b = pre[i][1];    if(b==v) {    //到这里，说明b是访问过的    if(visited[a]==0) {    visited[a]=1;    flag = DFS(a);    visited[a]=2;    if(!flag) return false;    }else if(visited[a]==1) return false;    }      }        return flag;    }        //迭代    //TC=O(m^n)    public boolean canFinish_it(int numCourses, int[][] prerequisites) {    n = numCourses;    pre = prerequisites;        if(n==0 || n==1) return true;    if(pre==null || pre.length<1) return true;        visited = new int[n];    m = pre.length;        Stack<Integer> stack = new Stack<Integer>();    for(int i=0; i<n; i++) {    if(visited[i]==0) {    stack.push(i);    visited[i] = 1;    }    int v;    while(!stack.isEmpty()) {                v = stack.peek();//revise    int j=0;                int flag=0;    for(; j<m; j++) {//遍历边的时候，可以优化，将已经访问过的边标记    int a = pre[j][0];    int b = pre[j][1];    if(v==b) {    if(visited[a]==0) {                            flag=1;                            stack.push(a);            visited[a]=1;    break;    }else if(visited[a]==1) return false;    }    }    //revise    if(flag==0) {    v = stack.pop();    visited[v] = 2;     }    }    }    return true;    }        //方法二：根据拓扑排序，实质为BFS策略        public boolean canFinish_1(int numCourses, int[][] prerequisites) {    //空间换时间    n = numCourses;    pre = prerequisites;        m = pre.length;        int[] nums = new int[n];    for(int i=0; i<m; i++) {    nums[pre[i][1]]++;    }    Queue<Integer> q = new LinkedList<Integer>();    int count=0;    for(int i=0; i<n; i++) {    if(nums[i]==0) {    q.add(i);    count++;    }         }    while(!q.isEmpty()) {    int t = q.poll();    for(int i=0; i<m; i++) {    if(t==pre[i][0]) {    int k = pre[i][1];    nums[k]--;    if(nums[k]==0) {    q.add(k);    count++;    }    }    }    }    return count==n;    }        public static void main(String[] args) {    M207_Course_Schedule m = new M207_Course_Schedule();        m.canFinish_it(3, new int[][] {{1,0},{2,0},{0,2}});    }    }