题解 | #字符串合并处理#

1. 字符串合并，直接string类型相加即可
2. 排序，和很多人思路一样，利用multiset容器的有序性的，装进去在取出来
3. 二进制倒转，没想到用移位运算解决（这也是个好思路），就把字符串每个字符拆分，利用8421法手动转换为二进制存放在int[4]数组，然后倒序，在运算变为10进制，在转换为16进制的字符

#include<iostream>
#include<set>
using namespace std;

struct rule{
    bool operator()(char a,char b){
        return (a<b);
    }
};

string convertnum(string str)
{
    int temp;
    int num_10;
    for(int i=0 ; i<str.length();i++){
        if(str[i]>='0'&&str[i]<='9'){
            num_10 = str[i] - '0';
        }else if(str[i]>='a'&&str[i]<='f'){
            num_10 = str[i] - 'a' + 10;
        }else if(str[i]>='A'&&str[i]<='F'){
            num_10 = str[i] - 'A' + 10;
        }else{
            continue;
        }
        
        int num_2[4];
        num_2[3] = num_10/8;
        num_10 = num_10 % 8;
        num_2[2] = num_10/4;
        num_10 = num_10 % 4;
        num_2[1] = num_10/2;
        num_10 = num_10 % 2;
        num_2[0] = num_10/1;
        num_10 = num_2[3] + num_2[2]*2 + num_2[1]*4 + num_2[0]*8;
        if(num_10>=0&&num_10<=9){
            str[i] = num_10 + '0';
        }else if(num_10>9&&num_10<16){
            str[i] = num_10 + 'A' - 10;
        }else{
            cout<<"num_10 error! : "<<num_10 <<endl;
        }
    }
    return str;
}

int main()
{
    string str1,str2;
    multiset<char,rule>strset0;
    multiset<char,rule>strset1;
    string str;
    cin>>str1>>str2;
    str = str1 + str2;
    for(int i=0 ;i<str.length();i=i+2){
        strset0.insert(str[i]);
    }

    for(int i=1 ;i<str.length();i=i+2){
        strset1.insert(str[i]);
    }
    
    int n = 0;
    auto it0 = strset0.begin();
    auto it1 = strset1.begin();
    for(;it0!=strset0.end()&&it1!=strset1.end();it0++,it1++){
            str[n] = *it0;
            n++;
            str[n] = *it1;
            n++;
    }
    if(it0!=strset0.end()){
        str[n] = *it0;
    }else if(it1!=strset1.end()){
        str[n] = *it1;
    }

    str = convertnum(str);
    cout<<str;
    
}