hdoj1004(查找众多字符串中个数最多的字符串)

Let the Balloon Rise.

最近开始刷hdoj，想通过写博客做做笔记，记录写过代码。

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

分析：

本题的目的是将众多字符串中的出现个数最多的字符串找到并输出。

1.在用c/c++编写的时候，可以使用字符型的二维数组存字符串。数组一般声明得稍大些。

2.思路是计算每一个字符串s[j]后面的与s[j]相同的字符串的个数a[j]，若a[j]>max,则max=a[j];flag=j，最后flag中存储的就是个数最多的字符串第一次出现的下标。

3.用int strcmp(char[],char[])==0?来判断字符串是否相同。

下面是AC代码：

#include<stdio.h>
#include<string.h>
#define maxn 1005     //N<=1000,数组一般声明得稍大些，所以我设1005
int main(){
	int n,flag,max;    //max用来存当前颜色最多的字符串个数，flag存其下标
	int a[maxn];       //用来存放每个字符串后面的与其相同的字符串个数
	char s[maxn][16];
	while(scanf("%d",&n)!=EOF&&n){     
		max=0;
		flag=0;
		for(int i=0;i<n;i++)    //输入字符串
			scanf("%s",s[i]);
		for(int j=0;j<n;j++){    //比较两个字符串，如果相同，则对应a[j]++
			a[j]=0;
			for(int k=j+1;k<n;k++)    
				if(strcmp(s[j],s[k])==0)
					a[j]++;
			if(a[j]>max){
				max=a[j];
				flag=j;
			}
		}
		printf("%s\n",s[flag]);    //输出最大个数的字符串
	}
	return 0;
}

最后，我有个疑问，就是假如输入的字符串个数最多的不止一种，该怎么办？例如

red

pink

black

red

pink

我的处理方法是输出先达到max的字符串，也就是靠前的，即“red"。如果大家有别的见解，欢迎提出来给大家分享分享。