编码-腾讯机试

假定一种编码的编码范围是a-y的25个字母，从一位到四位的编码，如果我们把该编码按字典序排序，形成一个数组如下：
a，aa，aaa，aaaa，aaab，aaac，... ...，b，ba，baa，baaa，baab，baac，... ...，yyyw，yyyx，yyyy；
其中a的Index为0，aa的Index为1，aaa的Index为2，以此类推。
编写一个函数，输出给定编码对应Index。如：baca --> 16331.

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

#define N3 1
#define N2 25
#define N1 (25 * 25)
#define N0 (25 * 25 * 25)

#define C3 N3
#define C2 (N2 + N3)
#define C1 (N1 + N2 + N3)
#define C0 (N0 + N1 + N2 + N3)

int main()
{
    char code[11][5] = {"a", "aa", "aaa", "aaaa", "ab", "aba", "b", "ba", "bac", "baca", "yyyy"};
    
    for(int i = 0; i < 11; i++){
        int index = 0;
        switch(strlen(code[i])){
            case 1: index = C0 * (code[i][0] - 'a'); break;
            case 2: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + 1; break;
            case 3: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + 2; break;
            case 4: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + C3 * (code[i][3] - 'a') + 3; break;
            default: break;
        }
        printf("code = %-4s, strlen =%2d, index =%7d\n", code[i], strlen(code[i]), index);
    }
    return 0;   
}

结果显示如下：

code = a   , strlen = 1, index =      0
code = aa  , strlen = 2, index =      1
code = aaa , strlen = 3, index =      2
code = aaaa, strlen = 4, index =      3
code = ab  , strlen = 2, index =    652
code = aba , strlen = 3, index =    653
code = b   , strlen = 1, index =  16276
code = ba  , strlen = 2, index =  16277
code = bac , strlen = 3, index =  16330
code = baca, strlen = 4, index =  16331
code = yyyy, strlen = 4, index = 406899

完整截图：

编码-腾讯机试题.png

改进版代码：

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

#define N3 1
#define N2 25
#define N1 (25 * 25)
#define N0 (25 * 25 * 25)

#define C3 N3
#define C2 (N2 + N3)
#define C1 (N1 + N2 + N3)
#define C0 (N0 + N1 + N2 + N3)

int main()
{
    char code[11][5] = {"a", "aa", "aaa", "aaaa", "ab", "aba", "b", "ba", "bac", "baca", "yyyy"};
    
    for(int i = 0; i < 11; i++){
        int index = 0;
        switch(strlen(code[i])){
            case 4: index += C3 * (code[i][3] - 'a') + 1;
            case 3: index += C2 * (code[i][2] - 'a') + 1;
            case 2: index += C1 * (code[i][1] - 'a') + 1;
            case 1: index += C0 * (code[i][0] - 'a');
        /*
            case 1: index = C0 * (code[i][0] - 'a'); break;
            case 2: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + 1; break;
            case 3: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + 2; break;
            case 4: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + C3 * (code[i][3] - 'a') + 3; break;
        */
            default: break;
        }
        printf("code = %-4s, strlen =%2d, index =%7d\n", code[i], strlen(code[i]), index);
    }
    return 0;   
}

改进版完整截图：

改进版完整截图.png