两道BFS拓扑排序

Course Schedule

经典题。检验图中是否成环。

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        ArrayList[] graph = new ArrayList[numCourses]; // 每个index表示从哪个出发，index的arraylist包含所有到达的
        int[] in = new int[numCourses]; // 每个顶点的入度
        
        for (int i = 0; i < graph.length; ++i) {
            graph[i] = new ArrayList();
        }
        
        for (int[] p: prerequisites) {
            graph[p[1]].add(p[0]);
            in[p[0]]++;
        }
        
        Queue q = new LinkedList();
        // for (int i : in) {
        //     if (i == 0) q.offer(i);
        // }
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.offer(i);
        }
        
        while (!q.isEmpty()) {
            int pre = q.poll();
            for (int post : graph[pre]) {
                --in[post];
                if (in[post] == 0) q.offer(post);
            }
        }
        
        for (int i = 0; i < in.length; ++i) {
            if (in[i] != 0) return false;
        }
        return true;
    }
}

值得注意的是入度为0的顶点加入q的时候是:

// for (int i : in) {
//     if (i == 0) q.offer(i);
// }
  for (int i = 0; i < numCourses; ++i) {
       if (in[i] == 0) q.offer(i);
  }

是遍历所有顶点加入入度为0的顶点，并不是遍历in入度数组加入这些入度 -- 加入入度0并没有任何意义。这里注意一下。

Alien dictionary

根据已排序相邻的两个字符串构建Map>,同时对每个char保存入度。本质上就是多了一个char->set的集合。然后还是用q进行BFS拓扑排序。最后根据结果规模和顶点数判断是否成环

class Solution {
    public String alienOrder(String[] words) {
        Map> map = new HashMap();
        Map indegree = new HashMap();
        String res = "";
        if (words == null || words.length == 0) return res;
        for (String s : words) {
            for (char c : s.toCharArray()) {
                indegree.put(c, 0);
            }
        }
        
        
        for (int i = 0; i < words.length - 1; ++i) {
            String cur = words[i];
            String next = words[i+1];
            int len = Math.min(cur.length(), next.length());
            for (int j = 0; j < len; ++j) {
                char c1 = cur.charAt(j);
                char c2 = cur.charAt(j);
                if (c1 != c2) { //[c1, c2]有没有记录下来
                    Set set = new HashSet();
                    if (map.containsKey(c1)) set = map.get(c1);
                    if (!set.contains(c2)){
                        set.add(c2);
                        map.put(c1, set);
                        indegree.put(c2, indegree.get(c2)+1);
                    }
                     break;
                }
            }
        }
        
        Queue q = new LinkedList();
        for (char c: indegree.keySet()){
            if (indegree.get(c) == 0) q.offer(c);
        }
        
        while (!q.isEmpty()) {
            char c = q.poll();
            res += c;
            if (map.containsKey(c)) {
                for (char c1 : map.get(c)){
                    indegree.put(c1, indegree.get(c1)-1);
                    if (indegree.get(c1) == 0) q.offer(c1);
                }
            }
        }
        
        if (res.length() != indegree.size()) return ""; // non-DAG, loop exists; contradiction
        return res;
    }
}