HDOJ Let the Balloon Rise（java）

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 69974    Accepted Submission(s): 26018


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 WA代碼：利用兩層for迴圈統計，但是總是報錯，不知道錯在哪裡，還請大牛指點。。。

import java.util.Scanner;public class Main {        public static void main(String[] args) {                Scanner in = new Scanner(System.in);        while(in.hasNext()){            int n = in.nextInt();            if(n == 0)                break;            String[]  str = new String[n];            boolean[] flag = new boolean[n];            for(int i=0; i<n; i++){                str[i] = in.next();                flag[i] = false;            }            int count = 1;            int max = Integer.MIN_VALUE;            int maxIndex = 0;            for(int i=0; i<n-1; i++){                if(flag[i] == true)                    continue;                flag[i] = true;                for(int j=i+1; j<n; j++){                    if(str[i].trim().equals(str[j].trim())){                        count++;                        flag[j] = true;                    }                }                                if(count > max){                    maxIndex = i;                    max = count;                    count = 1;                }            }                        System.out.println(str[maxIndex]);                    }    }}

AC代碼：利用map集合類統計

import java.util.HashMap;import java.util.Iterator;import java.util.Map;import java.util.Scanner;public class Main {        public static void main(String[] args) {                Scanner in = new Scanner(System.in);        while(in.hasNext()){            int n = in.nextInt();            int number = 1;            if(n == 0)                break;            Map<String,Integer> map = new HashMap<String,Integer>();            for(int i=0; i<n; i++){                String str = in.next();                if(!map.containsKey(str)){                    map.put(str,number);                }else{                    int temp = map.get(str);                    map.put(str,++temp);                }            }            int max = Integer.MIN_VALUE;            String maxColor = "";            Iterator<?> it = map.keySet().iterator();            while(it.hasNext()){                String key = (String) it.next();                int value = map.get(key);                if(value > max){                    max = value;                    maxColor = key;                }            }            System.out.println(maxColor);                    }    }}