Hdu1004 Let the Balloon Rise

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 114532 Accepted Submission(s): 44832

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

题意：输入n个颜色，输出出现次数最多的颜色。

思路：我的方法是用List集合，三个list，分别记录 n个输入，颜色数目和出现次数，list2和list3对应，之后遍历list1，记录每种颜色出现的次数，然后找到最大的，输出。其实也可以把数据存在数组中，排序，找出最大出现次数，再遍历根据这个次数找寻并输出颜色，这里主要复习了一下List的使用，也熟悉了List中的set方法。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
         Scanner sc = new Scanner(System.in);
         while(sc.hasNext()){
        	 int n = sc.nextInt();
        	 if(n==0){
        		 break;
        	 }
        	 List<String> list1 = new ArrayList<String>();
        	 List<String> list2 = new ArrayList<String>();
        	 List<Integer> list3 = new ArrayList<Integer>();
        	 
        	 while(n-->0){
        		 String str = sc.next();
        		 list1.add(str);
        		 if(!list2.contains(str)){
        			 list2.add(str);
        		 }
        	 }
        	 for(int i=0;i<list2.size();i++){
        		 list3.add(0);
        	 }
        	 for(int i=0;i<list1.size();i++){
        		 for(int j=0;j<list2.size();j++){
        			if(list1.get(i).equals(list2.get(j))){
        				list3.set(j, list3.get(j)+1);
        				break;
        			}
        		 }
        	 }
        	 int max=0;
        	 int pos=0;
        	 for(int i=0;i<list3.size();i++){
        		 if(list3.get(i)>max){
        			 max=list3.get(i);
        			 pos=i;
        		 }
        	 }
        	 System.out.println(list2.get(pos));
         }
	}

}