The quick and dirty code I wrote was

#include
#include

class s {
private:
char indx;
char ondx;
char v[10];
public:
bool done;
unsigned long long count;
s(void) {
indx = '\0';
ondx = '\0';
done = false;
count= 0UL;
}
void put (char c) {
v[indx++] = c;
indx = indx % 10;
}
char get (void) {
char c = v[ondx++];
ondx = ondx %10;
return c;
}
char peek (void) {
return v[ondx];
}
bool isEmpty (void) {
return indx == ondx;
}
};

s context[100];

// the two functions "nextItem" and "doCount" call each other recursively.

bool nextItem (char& c, int level);

// count number of consecutive instances of v from level below
// return count as a character in 'c'; save v with msb set. and
// character which terminated count in context.
void doCount (char& c, char v, int level) {
bool r = false;
s& x = context[level];
c = '1';
x.put (v + 0X80);
while (!x.done) {
char t;
x.done = nextItem (t, level - 1);
if (t == v) {
c++;
}
else {
// count is complete so we need to put the terminating value
// in the buffer, value we are counting is already there
x.put (t);
break;
}
}
}

// return in 'c' the next character from the specified level
// signal done if that character is the last.
// there are two special cases:
// 1) level = 0, the single character '1' is returned and done is signalled
// 2) There are no characters held in the context (this must be the first entry)
// Otherwise there are one or two characters in the context. If the next character
// held in the context ha the msb set, the count has already been returned so the
// character should be returned. Otherwise the character is the terminating character
// from the last count and more repetitions (if any) must be counted, after which the
// count is returned. When the lower level has signalled done, then when the last
// character is returned also signal done.
// Count the number of characters returned and when the last character is returned and
// done s signalled, report the level and the total.

bool nextItem (char& c, int level) {
s& x = context[level];
bool r = false;
if (!x.isEmpty()) {
// more ready to output
char v = x.get();
c = v & 0X7F;
if (v & 0X80) {
r = x.isEmpty();
}
else {
// this is the next value and we need to count any more
doCount (c, v, level);
}
}
else if (level == 0) {
// at the lowest level the seed is a single '1'
c

Since the only requirement was to determine the length, it is not necessary to store the full string. A simple 100 level recursion that, at each level, returns the next digit in sequence suffices - it takes a long time to run, but does not need huge amounts of space. At each level above the first it is only necessary to store at most two digits - the digit of which you have just counted the repetitions, and the digit that broke the sequence. Each invocation at any level alternates between returning the count and returning the counted digit.