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 are to balloons floating around. But to tell your a secret, the judges ' favorite time is guessing the most popular. When the contest was over, they'll count the balloons of each color and find the result.

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

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 lower-case letters.

A test Case with N = 0 terminates the "input" and "This" test case is processed.

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

Sample Input

5 green red Blue Red Red 3 pink Orange Pink 0
Sample Output

Red Pink WA Code: The use of two-tier for loop statistics, but always error, do not know where the wrong, but also please Daniel pointing ...

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 Code: Using Map Collection class statistics

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);
 }

    }

}