content: package question0423_reconstruct_original_digits_from_english;
  
  /**
   * 时间复杂度是O(n)，其中n是字符串s的长度。空间复杂度是O(1)。
   *
   * 执行用时：6ms，击败82.35%。消耗内存：41.4MB，击败6.28%。
   */
  public class Solution {
      public String originalDigits(String s) {
          int[] countLetter = new int[26];
          for (int i = 0; i < s.length(); i++) {
              countLetter[s.charAt(i) - 'a']++;
          }
          int[] countNum = new int[10];
          // "zero"       z
          countNum[0] = countLetter['z' - 'a'];
          // "one"        o - z - w - u
          countNum[1] = countLetter['o' - 'a'] - countLetter['z' - 'a'] - countLetter['w' - 'a'] - countLetter['u' - 'a'];
          // "two"        w
          countNum[2] = countLetter['w' - 'a'];
          // "three"      r - z - u
          countNum[3] = countLetter['r' - 'a'] - countLetter['z' - 'a'] - countLetter['u' - 'a'];
          // "four"       u
          countNum[4] = countLetter['u' - 'a'];
          // "five"       f - u
          countNum[5] = countLetter['f' - 'a'] - countLetter['u' - 'a'];
          // "six"        x
          countNum[6] = countLetter['x' - 'a'];
          // "seven"      s - x
          countNum[7] = countLetter['s' - 'a'] - countLetter['x' - 'a'];
          // "eight"      g
          countNum[8] = countLetter['g' - 'a'];
          // "nine"       i - g - x - f + u
          countNum[9] = countLetter['i' - 'a'] + countLetter['u' - 'a'] - countLetter['g' - 'a'] - countLetter['x' - 'a'] - countLetter['f' - 'a'];
          StringBuilder sb = new StringBuilder();
          for (int i = 0; i < 10; i++) {
              for (int j = 0; j < countNum[i]; j++) {
                  sb.append(i);
              }
          }
          return sb.toString();
      }
  }