输出最多的颜色

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

解法1:

# include <stdio.h>
    # include <string.h>

    int main(){

    char st[1005][20];
    int i,j,k,n,len,t;

    while(scanf("%d",&n),n){
        for(i=0;i<n;i++){
            scanf("%s",st[i]);
        }

        k = 0;
        t = 0;
        for(i=0;i<n;i++){
            int j = i+1,cnt = 0 ;
            for(;j<n;j++){
                if(strcmp(st[i],st[j])==0){
                    cnt++;
                } 
            }

            if(cnt>t){
                t = cnt;
                k = i;
                cnt = 0;
            }
        }

        printf("%s\n",st[k]);
    }


    return 0;
    }

解法2（map）：

利用map(与java中treemap对应,,--- java的hashmap就是c++的unordered_map（无序）)：
    # include <iostream>
    # include <map>
    # include <string>

    using namespace std;

    int main(){

    int n,i,j;
    map<string,int> mapstr; 
    string str;

    int max = 0;
    string ch;
    while(cin>>n,n){
        max = 0;
        while(n--){
            cin>>str;
            if(mapstr[str]==0){
                mapstr[str] = 1;                
            }else{
                mapstr[str]++;
            }
            //只所以这样处理是因为map排序
            if(mapstr[str]>max){
                max = mapstr[str];
                ch = str;
            }   
        }
         cout<<ch<<endl;
         mapstr.clear();

    }


    return 0;
    }