leetcode 443. String Compression

下面反向遍历，还是正向好。

    void left(vector<char>& v, bool p(int)) {
        int max_index = v.size() - ;

        int del = -;
        int rel = -;
        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<char>& chars) {
        int size = chars.size();
        int point = size - ;
        int count = ;
        for (int i = point; i >= ; i--) {
            if (chars[i] == chars[i - ] && i > )
                count++;
            else if (count > ) {
                for (int j = count - ; j > ; j--)
                    chars[i + j] = ;
                string temp = to_string(count);
                for (int j = ; j < temp.size(); j++)
                    chars[i + j + ] = temp[j];
                count = ;
            }
        }
        left(chars, [](int v) {return v != ;});
        return count_if(chars.begin(), chars.end(), [](int v) {return v != ;});
    }

其他答案：

    int compress(vector<char>& chars) {
        int lo=;
        int cnt=;
        for(int i=; i<chars.size(); i++){
            cnt++;
            if(i==chars.size()-||chars[i]!=chars[i+]){
                chars[lo++]=chars[i];
                if(cnt>){
                    string nums=to_string(cnt);
                    for(int i=; i<nums.length(); i++){
                        chars[lo++]=nums[i];
                    }
                }
                cnt=;
            }
        }
        return lo;
    }