Huawei online programming question series-14-string connection longest path search

Problem Description:

1. The problem involves knowledge points

  • string type sort
  • Repeat
  • Priority queue

2. Solve by yourself

  • Solution 1: directly compare the size of characters according to the idea of sorting, or use string.compareTo for comparison
package com.chaoxiong.niuke.huawei;

import java.util.Scanner;
/**
 * Create by tianchaoxiong on 18-4-9.
 * // The dictionary sort of strings can be sorted directly
 */
public class HuaWei_14 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        String[] stringArr = new String[N];
        int stringArrIndex = 0;
        for (int i = 0; i < N; i++) {
            inSert(scanner.next(), stringArr, stringArrIndex);
            stringArrIndex++;
        }
//        Arrays.sort(stringArr);
        for (String each : stringArr)
            System.out.println(each);

        }

    private static void inSert(String key, String[] stringArr, int stringArrIndex) {
        if(stringArrIndex==0){
            // If it's a direct insert
            stringArr[stringArrIndex]=key;
        }else {
            // Insert by insertion sort
            int i;
            for(i=stringArrIndex-1;i>=0;i--){
                if(isSmall(key,stringArr[i])){
                    //move
                    stringArr[i+1] = stringArr[i];
                }else
                    break;
            }
            stringArr[i+1] = key;

        }
    }
    private static boolean isSmall(String a, String b) {
        // Returns true if dictionary sequence a is less than b
        char []aArr = a.toCharArray();
        char []bArr = b.toCharArray();
        int index = 0;
        while (index<aArr.length&&index<bArr.length){
            if(aArr[index]<bArr[index]){
                return true;
            }else {
                if(aArr[index]>bArr[index]){
                    return false;
                }else {
                    index++;
                }
            }
        }
        return aArr.length == index;
    }
}
  • Solution 2: store the data in a priority queue. Traverse the output
package com.chaoxiong.niuke.huawei;

import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
/**
 * Create by tianchaoxiong on 18-4-9.
 */
public class HuaWei_14_2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        scanner.nextLine();
        Queue<String>priorityQueue = new PriorityQueue<String>(N);
        for (int i = 0; i < N; i++) {
            priorityQueue.add(scanner.nextLine());
        }
        while (priorityQueue.peek()!=null)
            System.out.println(priorityQueue.poll());
    }
}

3. High quality answer

  • Use ArrayList to store data
  • Use Collections.sort(arrayList) to sort the data
import java.util.*;
public class Main{
    public static void main(String[]args){
        Scanner scan=new Scanner(System.in);
        ArrayList<String> set=new ArrayList<String>();
        int num=scan.hasNextLine()?Integer.parseInt(scan.nextLine()):0;
        while(--num>=0&&scan.hasNextLine()){
             set.add(scan.nextLine());
        }
        Collections.sort(set);
        for(String str:set){
            System.out.println(str);
        }
    }
}

4. Summary of this question

string.compareTo:

// 123.compareTo("123") return 0;
// 123.compareTo("1234") return -1;
// 123.compareTo("12345") return -2;
// 1234.compareTo("12") return 2;
// 1234.compareTo("1237") return -3;
// 1234.compareTo("12375") return -3;
// 12375.compareTo("1233") return 4;

arrayList can directly put basic types of data, and then sort them by means of Collections.sort(obj)

Tags: Java less

Posted on Thu, 19 Mar 2020 14:25:03 -0400 by antonbrk