下面反向遍历，还是正向好。

void left(vector
     
      & v, bool p(int)) {        int max_index = v.size() - 1;        int del = -1;        int rel = -1;        while (del < max_index) {            while (p(v[del]) && del < max_index)                del++;            if (del >= max_index)                break;            if (rel < del)                rel = del;            while (!p(v[rel]) && rel <= max_index)                rel++;            if (rel > max_index)                break;            swap(v[del], v[rel]);            del++;        }    }    int compress(vector
      
       & chars) {        int size = chars.size();        int point = size - 1;        int count = 1;        for (int i = point; i >= 0; i--) {            if (chars[i] == chars[i - 1] && i > 0)                count++;            else if (count > 1) {                for (int j = count - 1; j > 0; j--)                    chars[i + j] = 2;                string temp = to_string(count);                for (int j = 0; j < temp.size(); j++)                    chars[i + j + 1] = temp[j];                count = 1;            }        }        left(chars, [](int v) {
   return v != 2;});        return count_if(chars.begin(), chars.end(), [](int v) {
   return v != 2;});    }
      
     

 

其他答案：

int compress(vector
     
      & chars) {        int lo=0;        int cnt=0;        for(int i=0; i
      
       1){                    string nums=to_string(cnt);                    for(int i=0; i