感觉华为的编程题考查的不是人的编程能力啊。。。

模块依赖关系分析    点击链接看源码

public class NIOTest { public static void main(String[] args) { Scanner sc = new Scanner(System.in); ArrayList<Node> res = new ArrayList<>(); HashMap<String, Node> relation = new HashMap<>(); boolean flag = true; while (sc.hasNext()) { String str = sc.nextLine(); parse(str, relation); if (str.charAt(str.length() - 1) != ',') break; } while (!relation.isEmpty() && flag == true) { flag = false; Node node = null; for (Map.Entry<String, Node> entry : relation.entrySet()) { Node cur = entry.getValue(); if (cur.degree == 0) { node = cur; break; } } if (node != null) { for (Node temp : node.relation) { temp.degree--; } res.add(node); relation.remove(node.name); flag = true; } } for (Map.Entry<String, Node> entry : relation.entrySet()) { res.add(entry.getValue()); } Collections.sort(res, new Comparator<Node>() { public int compare(Node a, Node b) { return a.name.compareTo(b.name); } }); for (int i = 0; i < res.size() - 1; i++) { boolean b = res.get(i).degree == 0 ? false : true; System.out.println("{" + res.get(i).name + ", " + b + "},"); } boolean b = res.get(res.size() - 1).degree == 0 ? false : true; System.out.println("{" + res.get(res.size() - 1).name + ", " + b + "}"); } public static void parse(String str, HashMap<String, Node> relation) { String[] strs = str.split(","); String A = strs[0].substring(1, strs[0].length()); String B = strs[1].substring(1, strs[1].length() - 1); Node nodeA = relation.get(A); if (nodeA == null) { nodeA = new Node(A, null, 1); relation.put(A, nodeA); } else { nodeA.degree++; } Node nodeB = relation.get(B); if (nodeB == null) { nodeB = new Node(B, nodeA, 0); relation.put(B, nodeB); } else { nodeB.relation.add(nodeA); } } public static class Node { public String name; public LinkedList<Node> relation = new LinkedList<>(); public int degree; public Node (String name, Node relation, int degree) { this.name = name; if (relation != null)this.relation.add(relation); this.degree = degree; } } }

感觉华为考的是编程思想，还有格式，不是AC，估计和网易一样，一张一张的试卷筛选

问个问题 为啥我这次 就没有收到 笔试  是不是我春招华为挂了 就没机会了？？？

为什么你们可以不按照他规定的接口来写？

第二题本地正常的JAVA代码，贴上去编译不过去，改了Main了导包了，直接跪！

我66 0 33.、。是不是稳挂

最后2道都没有百分之100，唉

通过率多少你

public class NIOTest { public static void main(String[] args) { Scanner sc = new Scanner(System.in); ArrayList<Node> res = new ArrayList<>(); HashMap<String, Node> relation = new HashMap<>(); boolean flag = true; while (sc.hasNext()) { String str = sc.nextLine(); parse(str, relation); if (str.charAt(str.length() - 1) != ',') break; } while (!relation.isEmpty() && flag == true) { flag = false; Node node = null; for (Map.Entry<String, Node> entry : relation.entrySet()) { Node cur = entry.getValue(); if (cur.degree == 0) { node = cur; break; } } if (node != null) { for (Node temp : node.relation) { temp.degree--; } res.add(node); relation.remove(node.name); flag = true; } } for (Map.Entry<String, Node> entry : relation.entrySet()) { res.add(entry.getValue()); } Collections.sort(res, new Comparator<Node>() { public int compare(Node a, Node b) { return a.name.compareTo(b.name); } }); for (int i = 0; i < res.size() - 1; i++) { boolean b = res.get(i).degree == 0 ? false : true; System.out.println("{" + res.get(i).name + ", " + b + "},"); } boolean b = res.get(res.size() - 1).degree == 0 ? false : true; System.out.println("{" + res.get(res.size() - 1).name + ", " + b + "}"); } public static void parse(String str, HashMap<String, Node> relation) { String[] strs = str.split(","); String A = strs[0].substring(1, strs[0].length()); String B = strs[1].substring(1, strs[1].length() - 1); Node nodeB = relation.get(B); if (nodeB == null) { nodeB = new Node(B, null, 1); relation.put(B, nodeB); } else { nodeB.degree++; } Node nodeA = relation.get(A); if (nodeA == null) { nodeA = new Node(A, nodeB, 0); relation.put(A, nodeA); } else { nodeA.relation.add(nodeB); } } public static class Node { public String name; public LinkedList<Node> relation = new LinkedList<>(); public int degree; public Node (String name, Node relation, int degree) { this.name = name; if (relation != null)this.relation.add(relation); this.degree = degree; } } }

第二题，很简单的拓扑排序，为了表示形式，愣是花了太长时间