Move-to-front(MTF) and Run-lenght encoding(RLE) algorithms

　　mtf算法（關於該算法：https://www2.cs.duke.edu/csed/algoprobs/beta/bw1.html）：

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

void mtf_encode(const char *s, unsigned len, int *code)
{
    int pos = 0, num, trace = 0;
    unsigned i, j, k;
    int is_hav[26], stack[26];
	
    memset(is_hav, 0, 26);
	
    while (*s)
    {
        if (!is_hav[*s - 'a'])
        {
            stack[pos] = *s - 'a' + 1;
            is_hav[*s - 'a'] = 1;
            ++pos;
            code[trace++] = pos;
        }
        else
        {
            j = 0;
            while (stack[j] != *s - 'a' + 1) j++;
            code[trace++] = j + 1, num = stack[j];
            while (j > 0) stack[j] = stack[j - 1], j--;
            stack[j] = num;
        }
        s++;
    }
}

void print_mtf_encode(int*code, int len)
{
    unsigned i;
    printf("[");
    for (i = 0; i < len; i++)
    {
        printf("%d", code[i]);
        if (i != len - 1)
            printf(", ");
    }
    printf("]\n");
}

int main()
{
    char s[] = "abcabcaaaaaaaab";
    int len = strlen(s);
    int *code = (int *)malloc(sizeof(int) * len);
    mtf_encode(s, len, code);
    print_mtf_encode(code, len);
    free(code);
    return 0;
}

　　上面算法不是錯的，根據題意和不一樣的思考角度能夠編寫不一樣的mtf算法，按字典序排序後的mtf算法（只統計出現的字符）：

void mtf_encode(const char *s, unsigned len, int *code)
{
    int num, trace = 0;
    unsigned i, pos = 0;
    int is_hav[26], stack[26];
	
    memset(is_hav, 0, 26);
	
    for (i = 0; i < len; i++)
	if (!is_hav[s[i] - 'a'])
            is_hav[s[i] - 'a'] = s[i] - 'a' + 1;
	
    for (i = 0; i < 26; i++)
	if (is_hav[i] != 0)
	    stack[pos++] = is_hav[i];

    while (*s)
    {
        pos = 0;
        while (stack[pos] != *s++ - 'a' + 1) pos++;
        code[trace++] = pos + 1, num = stack[j];
        while (pos > 0) stack[pos] = stack[pos - 1], pos--;
        stack[pos] = num;
    }
}

　　rle算法（瞭解該算法：https://en.wikipedia.org/wiki/Run-length_encoding）：

const char*digits = "0123456789";

void comp_bit(char *rle, int *i, int num)
{
    if (num > 9)
        comp_bit(rle, i, num / 10);
    rle[(*i)++] = digits[num%10];
}

void rle_encode(const char *s, char *rle)
{
    unsigned i = 0, j = 1;
    rle[i++] = *s++;
    while (*s)
    {
        if (rle[i - 1] != *s)
        {
            comp_bit(rle, &i, j);
            rle[i++] = *s, j = 1;
        }
        else j++;
        s++;
    }
    comp_bit(rle, &i, j);
}