17.优化算法之解决拓扑排序4

0.基础 

1.课程表1 

207. 课程表 - 力扣（LeetCode）

class Solution {
    public boolean canFinish(int n, int[][] p) {
        // 1. 准备⼯作
        int[] in = new int[n]; // 统计每⼀个顶点的⼊度
        Map<Integer, List<Integer>> edges = new HashMap<>(); // 邻接表存图
        // 2. 建图
        for (int i = 0; i < p.length; i++) {
            int a = p[i][0], b = p[i][1]; // b -> a
            if (!edges.containsKey(b)) {
                edges.put(b, new ArrayList<>());
            }
            edges.get(b).add(a);
            in[a]++;
        }
        // 3. 拓扑排序
        Queue<Integer> q = new LinkedList<>();
        // (1) 先把⼊度为 0 的点，加⼊到队列中
        for (int i = 0; i < n; i++) {
            if (in[i] == 0)
                q.add(i);
        }
        // (2) bfs
        while (!q.isEmpty()) {
            int t = q.poll();
            for (int a : edges.getOrDefault(t, new ArrayList<>())) {
                in[a]--;
                if (in[a] == 0)
                    q.add(a);
            }
        }
        // 4. 判断是否有环
        for (int x : in)
            if (x != 0)
                return false;

        return true;
    }
}

2.课程表2

210. 课程表 II - 力扣（LeetCode）

class Solution {
    public int[] findOrder(int n, int[][] prerequisites) {
        // 1. 准备⼯作
        int[] in = new int[n]; // 统计每个顶点的⼊度
        List<List<Integer>> edges = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            edges.add(new ArrayList<>());
        }
        // 2. 建图
        for (int i = 0; i < prerequisites.length; i++) {
            int a = prerequisites[i][0], b = prerequisites[i][1]; // b -> a
            edges.get(b).add(a);
            in[a]++;
        }
        // 3. 拓扑排序
        Queue<Integer> q = new LinkedList<>();
        int[] ret = new int[n];
        int index = 0;
        for (int i = 0; i < n; i++) {
            if (in[i] == 0)
                q.add(i);
        }
        while (!q.isEmpty()) {
            int t = q.poll();
            ret[index++] = t;
            for (int a : edges.get(t)) {
                in[a]--;
                if (in[a] == 0)
                    q.add(a);
            }
        }
        if (index == n)
            return ret;
        return new int[0];
    }
}

3.火星词典

LCR 114. 火星词典 - 力扣（LeetCode）

 

class Solution {
    Map<Character, Set<Character>> edges = new HashMap<>(); // 邻接表
    Map<Character, Integer> in = new HashMap<>(); // 统计每个节点的⼊度
    boolean check;

    public String alienOrder(String[] words) {
        // 1. 初始化⼊度哈希表 + 建图
        for (String s : words) {
            for (int i = 0; i < s.length(); i++) {
                char ch = s.charAt(i);
                in.put(ch, 0);
            }
        }
        int n = words.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                add(words[i], words[j]);
                if (check == true)
                    return "";
            }
        }
        // 2. 拓扑排序
        Queue<Character> q = new LinkedList<>();
        for (char ch : in.keySet()) {
            if (in.get(ch) == 0)
                q.add(ch);
        }
        StringBuffer ret = new StringBuffer();
        while (!q.isEmpty()) {
            char t = q.poll();
            ret.append(t);
            if (!edges.containsKey(t))
                continue;
            for (char ch : edges.get(t)) {
                in.put(ch, in.get(ch) - 1);
                if (in.get(ch) == 0)
                    q.add(ch);
            }
        }
        // 3. 判断
        for (char ch : in.keySet()) {
            if (in.get(ch) != 0)
                return "";
        }
        return ret.toString();
    }

    public void add(String s1, String s2) {
        int n = Math.min(s1.length(), s2.length());
        int i = 0;
        for (; i < n; i++) {
            char c1 = s1.charAt(i), c2 = s2.charAt(i);
            if (c1 != c2) {
                // c1 -> c2
                if (!edges.containsKey(c1)) {
                    edges.put(c1, new HashSet<>());
                }
                if (!edges.get(c1).contains(c2)) {
                    edges.get(c1).add(c2);
                    in.put(c2, in.get(c2) + 1);
                }
                break;
            }
        }
        if (i == s2.length() && i < s1.length())
            check = true;
    }
}