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 -> aif (!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) bfswhile (!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 -> aedges.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 -> c2if (!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;}
}