273. Integer to English Words

1. Question

Convert a non-negative integer num to its English words representation.

2. Examples

Example 1:

Input: num = 123
Output: "One Hundred Twenty Three"

Example 2:

Input: num = 12345
Output: "Twelve Thousand Three Hundred Forty Five"

Example 3:

Input: num = 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

Example 4:

Input: num = 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

3. Constraints

  • 0 <= num <= 231 - 1

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/integer-to-english-words 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

百单位需要额外注意,每三个数为一个大单位。所以可以剥离成一个单独的函数。

遍历模拟即可。

class Solution {

  static String[] small = {
    "Zero", "One", "Two", "Three", "Four", "Five",
    "Six", "Seven", "Eight", "Nine", "Ten", 
    "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen",
    "Sixteen", "Seventeen", "Eighteen", "Nineteen"
  };
  static String[] medium = {
    "", "", "Twenty", "Thirty", "Forty", "Fifty",
    "Sixty", "Seventy", "Eighty", "Ninety"
  };
  static String[] large = {
    "", "Thousand", "Million", "Billion"
  };

  private StringBuilder num2Str(int x) {
    StringBuilder sb = new StringBuilder();
    if (x >= 100) {
      sb.append(small[x / 100]).append(" Hundred").append(" ");
      x %= 100;
    }
    if(x >= 20) {
      sb.append(medium[x / 10]).append(" ");
      x %= 10;
    }
    if (x != 0) {
      sb.append(small[x]).append(" ");
    }
    return sb;
  }

  public String numberToWords(int num) {
    if (num == 0) {
      return small[num];
    }
    StringBuffer sb = new StringBuffer();
    int count = 0;
    while (num != 0) {
      int k = num % 1000;

      if (k != 0) {
        StringBuilder s = num2Str(k);
        s.append(large[count]).append(" ");
        sb.insert(0, s);
      }

      count++;
      num /= 1000;
    }
    return sb.toString().trim();
  }
}
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