HDU-1004 JAVA

怼烎@ 2023-06-27 11:20 1阅读 0赞

Let the Balloon Rise  
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 173125 Accepted Submission(s): 68950

Problem Description  
Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges’ favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

Input  
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) – the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

Output  
For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5  
green  
red  
blue  
red  
red  
3  
pink  
orange  
pink  
0

Sample Output

red  
pink

Author  
WU, Jiazhi

Source  
ZJCPC2004

**这里用到LinkedList用来存数据，在用一个数据来存字符串出现的个数**

import java.util.LinkedList;
    import java.util.Scanner;
    
    public class Main{
        public static void main(String args[]) {
            Scanner sc=new Scanner(System.in);        
            while(sc.hasNext()){
                int count=sc.nextInt();
                int mun=count;
                if(count!=0){
                    int max[]=new int[1000];
                    LinkedList<String> linkedList=new LinkedList<String>();
                    int op=0;
                    while(count-->0){    
                        String color=sc.next();
                        if(linkedList.contains(color) == false){
                            linkedList.add(op, color);
                            max[op]=1;op++;
                            
                        }
                        else{
                            max[linkedList.indexOf(color)]+=1;
                        }                    
                    }                
                    int maxx=0;
                    int index=0;
                    for(int m=0;m<mun;m++){
                        if(max[m]>maxx){
                            maxx=max[m];
                            index=m;
                        }
                    }
                    System.out.println(linkedList.get(index));
                }else{
                    break;
                }            
                
            }
            
        }
    }