【算法学习笔记】90.字符串处理 计数 SJTU OJ 1279. 打电话

字符串水题， 但是最后一步关于计算重复度的还是很好玩的地方。

另外就是注意代码简洁性、可读性的练习。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

char getNumber(char c){
    if(c=='-' or c=='Q' or c=='Z')
        return '-';
    if(c>='0' and c<='9')
        return c;
    if(c=='S')
        return '7';
    if(c=='V')
        return '8';
    if(c=='Y')
        return '9';
    if(c>='A' and c<='Z')
        return (c-'A')/3 + 2 + '0';
    return '-';
}

struct PhoneNumber
{
    string s;
    void format(string tmp){
        s="";
        for (int i = 0; i < tmp.length(); ++i)
        {
            char t = getNumber(tmp[i]);
            if(t!='-')
                s += t;
        }
    }
    void print(){
        for (int i = 0; i < 3; ++i)
        {
            cout<<s[i];
        }
        cout<<"-";
        for (int i = 3; i < s.length(); ++i)
        {
            cout<<s[i];
        }
    }
};

int n;
void init(){
    cin>>n;
}
PhoneNumber book[100000];
bool cmp_pn(const PhoneNumber& a, const PhoneNumber& b){
    return a.s < b.s;
}
void build(){
    string tmp;
    for (int i = 0; i < n; ++i)
    {
        cin>>tmp;
        book[i].format(tmp);
    }
    sort(book,book+n,cmp_pn);
}

void output(){ 
    int cur = 0;
    int cur_dp = 1;
    bool no_dup = true;
    for (int i = 1; i < n; ++i)
    {
        if(book[i].s == book[cur].s){
            cur_dp++;
            if(cur_dp>=2)
                no_dup = false;
        }else{
            if(cur_dp>=2){
                book[cur].print();
                cout<<" "<<cur_dp<<endl;
            }
            cur_dp = 1;
            cur = i;
        }
    }
    if(no_dup)
        cout<<"No duplicates."<<endl;
}
int main(int argc, char const *argv[])
{
    init();
    build();
    output();
    return 0;
}