字符串合并处理：C语言解法

#include<stdio.h>
int cmp(char *a, char *b){
    return (int)(*a-*b);
}
char reverse(int num){
    int re_num = 0;
    char re_char = 0;
    int bit[4] = {0};
    for(int i=0; i<4; i++){
        bit[i] = num%2;
        num=num/2;
    }
    for(int i=0; i<4; i++){
        //对二进制数进行翻转并计算
        re_num += bit[3-i]*(int)pow(2,i);
    }
    if(re_num>=0 && re_num<=9)
        re_char = re_num + '0';
    else if(re_num>=10 && re_num<=15)
        re_char = re_num-10+'A';
    return re_char;
}
void convert(char *str){
    int len = strlen(str);
    int num = len/2;
    char str0[num];
    char str1[len-num-1];
    memset(str0, 0, num);
    memset(str1, 0, len-num-1);
    //第一步：将输入的两个字符串str1和str2进行前后合并
    for(int i=0,j=0; i<len; i++){
        if(str[i] == ' ')
            j++;
        str[i] = str[j++];
    }
    //第二步：对合并后的字符串进行排序
    for(int i=0,j=0; i<len-1; i++){
        if(i%2)
            str1[j++] = str[i];
        else
            str0[j] = str[i];
    }
    qsort(str0, num, sizeof(char), cmp);
    qsort(str1, len-num-1, sizeof(char),cmp);//排序数组的长度要给精确的值
    for(int i=0,j=0; i<len-1; i++){
        if(i%2)
            str[i] = str1[j++];
        else
            str[i] = str0[j];
    }
    //第三步：对排序后的字符串中的'0'~'9'、'A'~'F'和'a'~'f'字符，需要进行转换操作
    for(int i=0; i<len-1; i++){
        int temp = 16;
        if(str[i]>='A' && str[i]<='F'){
            temp = str[i]-'A'+10;
        }else if(str[i]>='a' && str[i]<='f'){
            temp = str[i]-'a'+10;
        }else if(str[i]>='0' && str[i]<='9'){
            temp = str[i]-'0';
        }
        if(temp != 16)
            str[i] = reverse(temp);
    }
    printf("%s\n",str);
}
int main(){
    char str[200] = {""};
    while(gets(&str)){
        convert(str);
    }
}