# 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"

## Thoughts

英语中都是每三位一切，如23,000。因此从个位开始往上, 每次/1000进行后对num%1000进行dfs后和对应的THOUSANDS写法拼接。DFS内部判断num是否等于0，小于20, 100或其它, 把高位先摘出来, 再递归处理低位, 低位可以通过取mod直接得到。

## Code

```cpp
/*
 * @lc app=leetcode id=273 lang=cpp
 *
 * [273] Integer to English Words
 */

// @lc code=start
class Solution
{
public:
    const vector<string> LESS_THAN_20{"", "One", "Two", "Three", "Four", "Five", "Six", "Seven",
                                 "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen",
                                 "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    const vector<string> TENS{"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    const vector<string> THOUSANDS{"", "Thousand", "Million", "Billion"};

    string dfs(const int num)
    {
        if (num == 0)
            return "";
        else if (num < 20)
            return LESS_THAN_20[num] + " ";
        else if (num < 100)
            return TENS[num / 10] + " " + dfs(num % 10);
        else
            return LESS_THAN_20[num / 100] + " Hundred " + dfs(num % 100);
    }
    string numberToWords(int num)
    {
        if (num == 0)
            return "Zero";
        string res;
        for (int i = 0; num > 0; num /= 1000, ++i)
        {
            if (num % 1000 != 0)
            {
                res = dfs(num % 1000) + THOUSANDS[i] + " " + res;
            }
        }
        const auto trim = [](string &s) {
            while (s.compare(0, 1, " ") == 0)
                s.erase(s.begin());
            while (s.size() > 0 && s.compare(s.size() - 1, 1, " ") == 0)
                s.erase(s.end() - 1);
        };
        trim(res);
        return res;
    }
};
// @lc code=end

```

```java
class Solution {
    private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven",
                                             "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen",
                                             "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen","Nineteen"};
    private final String[] TENS = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};

    private String helper(int num) {
        if (num == 0) {
            return "";
        } else if (num < 20) {
            return LESS_THAN_20[num] + " ";
        } else if (num < 100) {
            return TENS[num / 10] + " " + helper(num % 10);
        } else {
            return LESS_THAN_20[num / 100] + " Hundred " + helper(num % 100);
        }
    }

    public String numberToWords(int num) {
        if (num == 0) {
            return "Zero";
        }
        String res = "";
        int i = 0;
        while (num > 0) {
            if (num % 1000 != 0) {
                res = helper(num % 1000) + THOUSANDS[i] + " " + res;
            }
            num /= 1000;
            i++;
        }

        return res.trim();
    }
}
```

## Analysis

时间复杂度O(N), N为位数.