Alibaba telephone interview-Alibaba interview to solve two conflicts

Alibaba telephone interview-Alibaba interview to solve two conflicts

Situation: Party A and Party B handle some matters, each of which has a start time and a deadline. However, the agent issues of Party A and Party B may conflict. If there are other agent issues in a code issue, it is considered as a conflict. It is required to merge the agent items of Party A and Party B. There cannot be conflicts.

Solution:

I often use java to simulate this problem.

Define the event handling:

public class Node{private int startTime;private int endTime;Node cNode;}

The agent items of Party A and Party B are already ordered linked lists.

Initialize agent:

Node nodeA = new Node (1, 2); // initialize Node AnodeA. addNode (new Node (2, 6); nodeA. addNode (new Node (3, 7); nodeA. addNode (new Node (4, 9); nodeA. addNode (new Node (7, 12); nodeA. addNode (new Node (16, 19 ));

Initialize the agent of Party B:

Node nodeB = new Node(2, 4);nodeB.addNode(new Node(3, 6));nodeB.addNode(new Node(7, 10));nodeB.addNode(new Node(9, 15));nodeB.addNode(new Node(18, 20));

Note: There may be conflicts between the agent and the agent after the merger.

Based on this, we will directly merge the agent items of Party A and Party B, and then delete the conflicting items from the agent items after the merger.

PrintNode (nodeA); printNode (nodeB); Node mergeNode = mergeWithNoSort (nodeA, nodeB); // merge two nodes into printNode (mergeNode); Node deleteNode = deleteNode (mergeNode ); // Delete the conflicting node printNode (deleteNode );

Running result

Merge two nodes:

Public static Node mergeWithNoSort (Node nodeA, Node nodeB) {Node tempNode = nodeA; // merge nodeA and nodeB // USE insert and delete to insert nodeB into nodeA Node insertNode = nodeB; while (tempNode! = Null) {if (tempNode. cNode = null) {if (insertNode. startTime <= tempNode. startTime) {Node tempInsert = insertNode; insertNode. cNode = tempNode; tempNode = insertNode; insertNode = tempInsert; break;} else {tempNode. cNode = insertNode; return nodeA;} else {if (insertNode. startTime <= tempNode. startTime) {Node tempInsert = insertNode; insertNode. cNode = tempNode; tempNode = insertNode; insertNode = tempInsert; break;} else if (insertNode. startTime> tempNode. startTime & insertNode. startTime <= tempNode. cNode. startTime) {Node tempInsert = insertNode. cNode; insertNode. cNode = tempNode. cNode; tempNode. cNode = insertNode; // The clever use of nodeB is already in order. Write less the cycle insertNode = tempInsert;} else {tempNode = tempNode. cNode; }}return nodeA ;}

Function for deleting duplicate nodes:

// Delete the public static Node deleteNode (Node node) {Node tempNode = node; while (tempNode! = Null & tempNode. cNode! = Null) {if (tempNode. cNode. startTime <tempNode. endTime) // if a conflict exists, delete the Node {Node temp = tempNode. cNode; tempNode. cNode = temp. cNode; continue;} tempNode = tempNode. cNode;} return node ;}

Print node functions:

// Print a public static void printNode (Node node) {Node tempNode = node; while (tempNode! = Null) {System. out. print ("(" + tempNode. startTime + "," + tempNode. endTime + ")->"); tempNode = tempNode. cNode;} System. out. println ();}

Add a node to the last node of the node:

// Add a public void addNode (Node node) {Node tempNode = this; while (tempNode. cNode! = Null) {tempNode = tempNode. cNode;} tempNode. cNode = node ;}

In fact, it is not too difficult logically to analyze it carefully. For a person who hasn't written algorithms for a long time, when the interview is successful, the interview is rejected because of the tragedy. It seems that before looking for a job, you must write more code, review the data structure more.