（新人发帖）字节跳动笔试题第三题发工资个人Java解法


import java.util.*;
public class Main {

    public static void main(String[] args) {
      Scanner sc=new Scanner(System.in);
      int n=sc.nextInt();
      int[] group=new int[n];
      
      for(int i=0;i<n;i  ) {
          group[i]=sc.nextInt();
      }
      
      int min=10000, startPoint=0, salary=100, base1=100, base2=100, temp=0;
      for(int i=0;i<n;i  ) {
          if (group[i]<min) {
              min=group[i];
              startPoint=i;
          }
      }
      temp=startPoint;
      
    
      while(startPoint>1){
          if(group[startPoint-1]<group[startPoint]&&group[startPoint-2]>group[startPoint-1]) {
              base1=100; 
          }
          if(group[startPoint-1]<group[startPoint]&&group[startPoint-2]<group[startPoint-1]) {
              base1-=100; 
          }
          if(group[startPoint-1]>group[startPoint]){
              base1 =100;
              
          }
          salary =base1;
                    startPoint--; } while(startPoint==1) {
                     if(group[startPoint-1]>group[startPoint]) {
              base1 =100;
          }
          else {
              base1=100;
          }
          salary =base1;
          startPoint--;
         
      }
     
      startPoint=temp;
     
     while(startPoint<n-2) {
          if(group[startPoint 1]<group[startPoint]&&group[startPoint 2]>group[startPoint 1]) {
              base2=100;
          }
          if(group[startPoint 1]<group[startPoint]&&group[startPoint 2]<group[startPoint 1]) {
              base2-=100;
          }
          if(group[startPoint 1]>group[startPoint]){
              base2 =100;
          }
    
          salary =base2;
          startPoint  ;
         
      }
     while(startPoint==n-2) {
         if(group[startPoint 1]>group[startPoint]) {
             base2 =100;
         }
         else {
             base2=100;
         }
         salary =base2;
       startPoint  ;
   
     }
      System.out.println(salary);
    }

}

思路是先把所有人的工龄存到一个int数组，找到工龄最短的人的下标，然后往两边遍历。遇到工龄更大的人就此人工资加100，然后加进总；如果工龄比当前的人更小，则看是否比另一端的人更小（如果在边界就只需要考虑一侧），如果是就把工资重归100；其他情况工资则减少一百。因为两边遍历要分开考虑，所以初始化了base1和base2两个int来更新每个人的工资值。应该还有优化的空间。抛砖引玉和大佬们交流下哈#字节跳动##笔试题目#

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@不蛋定

西安邮电大学 Java

我当时思路和你一样，但没有考虑另一端的人工龄更小的问题，只有40通过。我再去试试你的思路。刚刚答完看见有位大佬的帖子，思路很清晰。试着按大佬的思路写了一下。 import java.util.*; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); TreeMap<Integer, ArrayList<Integer>> map = new TreeMap<>(); for (int i = 0; i < n; i++) { int score = scanner.nextInt(); if (!map.containsKey(score)) { map.put(score, new ArrayList<>()); } map.get(score).add(i); } int[] money = new int[n]; int[] temp = new int[n]; while (!map.isEmpty()) { int lowestKey = map.firstKey(); for (Integer i : map.get(lowestKey)) { if (i == 0) { temp[i] = money[i + 1] + 100; }else if (i == n - 1) { temp[i] = money[i - 1] + 100; } else { temp[i] = Math.max(money[i - 1], money[i + 1]) + 100; } } for (Integer i : map.get(lowestKey)) { money[i] = temp[i]; } map.remove(lowestKey); } int sum = 0; for (Integer i : money) { sum += i; } System.out.println(sum); } }

（新人发帖）字节跳动笔试题第三题发工资个人Java解法

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