leetcode-【hard】273. Integer to English Words

题目：

273. Integer to English Words

Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 1.

For example,

123 -> "One Hundred Twenty Three"
12345 -> "Twelve Thousand Three Hundred Forty Five"
1234567 -> "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

答案：

按着自己用英文阅读整数的顺序来就行：

1、这道题目不是难在思路上，而是难在考虑的情况比较多，比较多细节，一不小心就有小bug，需要思路清晰，逻辑清晰

2、英文中用来表示整数的单词并不多，分类：

1-9（个位数）,11-19（特殊的十位数），10-90（十位数），100（hundred），1000（thousand），1000000（million），1000000000（billion）

3、10这个十位数比较特别，只有在后两位完全为10时，才用ten

4、注意空格，后面没有数字了，就不能加空格了

5、每千位数跟后面的千位数（如果不为空，即0000000……）要有空格

6、注意不要乱加前缀空格和后缀空格

将上面的细节注意了就AC啦

代码：

#include <vector>
#include <string>

using std::vector;
using std::string;

string n2s[] = {"One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
string g2s[] = {"Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
string t2s[] = {"Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};

class Solution {
private:
    vector<string> ans;
    
public:
    string numberToWords(int num) {
        if(num == 0)
        {
            return "Zero";
        }
        
        unsigned int rem;
        unsigned int one;
        unsigned int two;
        unsigned int three;
        
        string perAns;
        ans.clear();
        
        do
        {
            //获取余数
            rem = num % 1000;
            num = num /1000;
            
            //存储每次计算的结果
            perAns.clear();
            perAns = "";
            
            //一千以内的数，取每一位
            one = rem % 10;
            two = (rem / 10) % 10;
            three = (rem / 100) % 10;
            
            if(three != 0)
            {
                perAns += n2s[three - 1];
                perAns += " Hundred";
            }
            
            //如果后两位都为0，那么后面就没有数了，也就不需要加空格
            if(perAns != "" && (one != 0 || two != 0))
            {
                perAns += " ";
            }
            
            if(two == 1)
            {
                //考虑最后一位是否为0的情况，因最后两位为10时，需要用“ten”
                //否则就是十位数
                if(one == 0)
                {
                    perAns += t2s[0];
                }else
                {
                    perAns += g2s[one - 1];
                }
            //考虑two不为0，不为1，则按一般规则去计算
            }else if(two != 0)
            {
                perAns += t2s[two - 1];
                
                if(one != 0)
                {
                    perAns += " ";//如果最后一位不为0，需要在其前面加空格
                    perAns += n2s[one - 1];
                }
            //考虑two为0的情况
            }else
            {
                if(one != 0)
                {
                    perAns += n2s[one - 1];
                }
            }
            
            //将结果存储到ans中，ans中的答案是以逆序形式存储了每个千位数
            ans.push_back(perAns);
        }while(num != 0);
        
        string result = "";
        unsigned int len = ans.size();
        //len最大长度为4，考虑每种位数就行，这里用的时候就会发现，合理使用goto，程序逻辑会很清晰
        switch(len)
        {
            case 4:goto three;break;
            case 3:goto two;break;
            case 2:goto one;break;
            case 1:goto zero;break;
        }
        
        three:
        if(ans[3] != "")
        {
            result += ans[3];
            result += " Billion";
            
            if(ans[2] != "" || ans[1] != "" || ans[0] != "")
            {
                result += " ";
            }
        }
        
        two:
        if(ans[2] != "")
        {
            result += ans[2];
            result += " Million";
            
            if(ans[1] != "" || ans[0] != "")
            {
                result += " ";
            }
        }
        
        one:
        if(ans[1] != "")
        {
            result += ans[1];
            result += " Thousand";
            
            if(ans[0] != "")
            {
                result += " ";
            }
        }
        
        zero:
        if(ans[0] != "")
        {
            result += ans[0];
        }
        
        return result;
    }
};

说明： 合理使用goto会取到很好的效果哦