具体思路：

个人想的是int的范围最大是10^9位，因此只需要考虑billion和million就可以；

将数字填充到12位，建立百位的转化，因此可以变为：
$3位+billion+3位+million+6位$；

其中6位又可以变成3位千+三位；

但是递归更好，这个题要二刷，恶心的一笔；

具体代码：

1.迭代版本：

class Solution {
public:
    string numberToWords(int num) {
        if(num==0)
            return "Zero";
        string s="";
        while(num!=0){
            s.push_back(num%10+'0');
            num/=10;
        }
        reverse(s.begin(), s.end());
        vector<string> temp=all_charge(s);
        return output(temp);
    }
    
    string output(vector<string> vec){
        string ret="";
        for(int i=0;i<vec.size();i++){
            if(i==0){
                ret+=vec[i];
            }else{
                ret+=" "+vec[i];
            }
        }
        return ret;
    }
    vector<string> all_charge(string s){
        while(s.size()!=12){
            s='0'+s;
        }
        vector<string>ret;
        string p1=s.substr(0,6);
        string pp1=p1.substr(0,3);
        string pp2=p1.substr(3,3);
        string p2=s.substr(6,6);
        vector<string> rp2=charge(p2);
        vector<string> rpp1=three_charge(pp1);
        vector<string> rpp2=three_charge(pp2);
        if(rpp1.size()!=0){
            ret.insert(ret.end(), rpp1.begin(),rpp1.end());
            ret.push_back("Billion");
        }
        if(rpp2.size()!=0)
            ret.insert(ret.end(), rpp2.begin(),rpp2.end());
        if(rpp2.size()!=0){
            ret.push_back("Million");
        }
        ret.insert(ret.end(),rp2.begin(),rp2.end());
        return ret;
    }

    vector<string> charge(string s){
        //最多六位，低三位为百，高三位为千；
        while(s.size()!=6){
            s='0'+s;
        }
        vector<string>ret;
        vector<string> p1=three_charge(s.substr(0,3));
        vector<string> p2=three_charge(s.substr(3,3));
        if(p1.size()!=0){
            ret.insert(ret.end(),p1.begin(),p1.end());
            ret.push_back("Thousand");
        }
        ret.insert(ret.end(),p2.begin(),p2.end());
        return ret;
    }

    vector<string> three_charge(string s){
        while(s.size()!=0&&s[0]=='0')
            s.erase(s.begin());
        vector<string> ret;
        if(s.size()==3){
            ret.push_back(low_digit[s[0]-'1']);
            ret.push_back("Hundred");
            s.erase(s.begin());
        }
        while(s.size()!=0&&s[0]=='0')
            s.erase(s.begin());
        if(s.size()==2){
            //去除前导零仍为两位；
            if(s[0]=='1'){
                ret.push_back(high_digit[s[1]-'0']);
            }else{
                ret.push_back(ten_digit[s[0]-'2']);
                if(s[1]!='0'){
                    ret.push_back(low_digit[s[1]-'1']);
                }
            }
        }else if(s.size()==1){
            //去除前导0为一位；
            ret.push_back(low_digit[s[0]-'1']);
        }
        return ret;
    }
private:
    vector<string> low_digit={"One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
    vector<string> high_digit={"Ten","Eleven","Twelve","Thirteen" ,"Fourteen" ,"Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
    vector<string> ten_digit={"Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
};

2.递归版本：

class Solution {
public:
    vector<string> singles = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};
    vector<string> teens = {"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    vector<string> tens = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};

    string comma(int n) {
        string ans = "";

        int hundred = n / 100;
        if (hundred) {
            ans += singles[hundred] + " Hundred ";
        }
        n = n % 100;
        if (n < 20 && n >= 10) {
            ans += teens[n%10] + " ";
            return ans;
        }
        if (n >= 20) {
            ans += tens[n/10] + " ";
        }
        n = n % 10;
        if (n == 0) return ans;

        ans += singles[n%10] + " ";
        return ans;
    }

    string recur(int n) {
        int billion = n / 1000000000;
        n %= 1000000000;
        int million = n / 1000000;
        n %= 1000000;
        int thousand = n / 1000;
        n %= 1000;
        string ans = "";

        if (billion) {
            ans += recur(billion) + "Billion ";
        }
        if (million) {
            ans += recur(million) + "Million ";
        }
        if (thousand) {
            ans += recur(thousand) + "Thousand ";
        }

        ans += comma(n);

        return ans;
    }

    string numberToWords(int num) {
        if (num == 0) return "Zero";
        string ans = recur(num);
        ans.pop_back();
        return ans;
    }
};