HDOJ 1004 java代码实现--引用相等要用 equals()Let the Balloon Rise

90 阅读 0 评论 60 点赞

我是靠谱客的博主霸气白开水，最近开发中收集的这篇文章主要介绍HDOJ 1004 java代码实现--引用相等要用 equals()Let the Balloon Rise，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

题目链接：点击打开链接

题目大意：求输入的n个字符串，数量最多的一共字符串。

本人第一次做，由于判断字符串是否相等，使用==判断，导致一直不对，后面查阅书本得知，引用需使用equals（）方法判断，方AC

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 119180 Accepted Submission(s): 46715

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




具体代码实现：


import java.util.*;
class Main{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
String str[]=new String[n];
int a[]=new int[n];
//该数组用于存放从该位置之后，一共有多少个字符串与该位置的字符串相同
for(int i=0;i<n;i++){
a[i]=0;
}
if(n==0){
break;
}
for(int i=0;i<n;i++){
str[i]=sc.next();
}
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){;
if(str[i].equals(str[j])){
//因为这里的str[]是引用/所以不能直接用==判断
a[i]++;
}
}
}
int t=0;
for(int i=0;i<n;i++){
if(a[i]>a[t]){
t=i;
//找到最大的那个就是最多
}
}
System.out.println(str[t]);
}
}
}