1. 程式人生 > >CodeWars 排隊買票的人手持25,50,100,判斷當前的零錢是否夠找零

CodeWars 排隊買票的人手持25,50,100,判斷當前的零錢是否夠找零

*Description:

The new “Avengers” movie has just been released! There are a lot of people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 dollars bill. A “Avengers” ticket costs 25 dollars.

Vasya is currently working as a clerk. He wants to sell a ticket to every single person in this line.

Can Vasya sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?

Return YES, if Vasya can sell a ticket to each person and give the change. Otherwise return NO.

Examples:
// * Java *

Line.Tickets(new int[] {25, 25, 50}) // => YES
Line.Tickets(new int []{25, 100})
// => NO. Vasya will not have enough money to give change to 100 dollars
*

public class Line {
    public static String Tickets(int[] peopleInLine){
        int bill25 = 0, bill50 = 0;
        for (int payment : peopleInLine){
            if(payment==25){
                bill25++;
            } else if(payment==50){
                bill25--;
                bill50++;
            } else
if(payment==100){ if(bill50>0){ bill50--; bill25--; } else{ bill25-=3; } } if(bill25<0 || bill50 <0){ return "NO"; } } return "YES"; } }

另一種方式:

public class Line {
  public static String Tickets(int[] peopleInLine)
  {
      int i, sum=0, change = 0;
      String a = "";
      for(i=0; i<peopleInLine.length; i++) {

          sum += 25;
          change = (peopleInLine[i] - 25);
          sum -= change;

          if(sum < change) {
            a = "NO";
          }
          else a = "YES";
      }
      return a;
  }
}