StringDup（理论AC）

Problem Statement for StringDup

Problem Statement

Create a class called StringDup. Given a string made up of ONLY letters and
digits, determine which character is repeated the most in the string ('A' is
different than 'a'). If there is a tie, the character which appears first in
the string (from left to right) should be returned.

Examples :

aaiicccnn = c
aabbccdd = a
ab2sbf2dj2skl = 2

Here is the method signature :

public char getMax(String input);

We will check to make sure that the input contains only letters and digits (no
punctuation marks or spaces).

Definition

Class:	StringDup
Method:	getMax
Parameters:	String
Returns:	char
Method signature:	char getMax(String param0)
(be sure your method is public)

---

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/*
* @Author: LyuCheng
* @Date:   2017-05-13 20:02:42
* @Last Modified by:   LyuCheng
* @Last Modified time: 2017-05-13 20:40:49
*/
/*
题意：给你一个只有字母和数字组成的字符串，让你输出出现次数最多的字符，如果最多数量的相同
    就输出最先出现的那个字符

思路：结构体加映射，结构体记录字符出现的顺序和出现的次数，然后按照次数为第一优先级，顺序
    第二优先级进行排序输出。
*/
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
struct node{
    int rank;
    int res;
    char str;
    node(){}
    node(int _rank,int _res,char _str){
        rank=_rank;
        res=_res;
        str=_str;
    }
    bool operator <(const node &other) const{
        if(res==other.res) return rank<other.rank;
        else return res>other.res;
    }
};
int pos=0;//字符的种类
node vis[65];
char str[1000005];
void init(){
    pos=0;
    for(int i=0;i<65;i++){
        vis[i].str='\0';
        vis[i].res=0;
        vis[i].rank=65;
    }
}
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%s",str)!=EOF){
        init();
        int n=strlen(str);
        for(int i=0;i<n;i++){
            if(str[i]>='0'&&str[i]<='9'){
                if(vis[str[i]-'0'].rank==65){
                    vis[str[i]-'0'].str=str[i];
                    vis[str[i]-'0'].rank=pos++;
                    vis[str[i]-'0'].res++;
                }else{
                    vis[str[i]-'0'].res++;
                }
            }else if(str[i]>='a'&&str[i]<='z'){
                if(vis[str[i]-'a'+10].rank==65){
                    vis[str[i]-'a'+10].str=str[i];
                    vis[str[i]-'a'+10].rank=pos++;
                    vis[str[i]-'a'+10].res++;
                }else{
                    vis[str[i]-'a'+10].res++;
                }
            }else{
                if(vis[str[i]-'A'+36].rank==65){
                    vis[str[i]-'A'+36].str=str[i];
                    vis[str[i]-'A'+36].rank=pos++;
                    vis[str[i]-'A'+36].res++;
                }else{
                    vis[str[i]-'A'+36].res++;
                }
            }
        }
        sort(vis,vis+61);
        printf("%c\n",vis[0].str);
    }
    return 0;
}