题目

Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.

Note:

  1. Input contains only lowercase English letters.
  2. Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as “abc” or “zerone” are not permitted.
  3. Input length is less than 50,000.

Example 1:

  1. Input: "owoztneoer"
  3. Output: "012"

Example 2:

  1. Input: "fviefuro"
  3. Output: "45"

题意

将一个数字字符串中的每一个数字转化为英文，并打乱所有字符顺序，要求还原出数字字符串。

思路

找规律题。依照一定的顺序，每一个数字的英文中都有一个唯一的字符能代表这个英文，如’z’只能代表”zero”。可以发现有以下规律：

  1. 剩余数字：[“zero”, “one”, “two”, “three”, “four”, “five”, “six”, “seven”, “eight”, “nine”]，其中具有特征字符的有：
    (‘z’, “zero”), (‘w’, “two”), (‘u’, “four”), (‘x’, “six”), (‘g’, “eight”)
  2. 剩余数字：[“one”, “three”, “five”, “seven”, “nine”]，其中具有特征字符的有：
    (‘o’, “one”), (‘h’, “three”), (‘f’, “five”)
  3. 剩余数字：[“seven”, “nine”]，其中具有特征字符的有：
    (‘s’, “seven”), (‘i’, “nine”)

按照上述顺序转化即可得到答案。

代码实现

Java

  1. class Solution {
  2.     public String originalDigits(String s) {
  3.         StringBuilder sb = new StringBuilder();
  4.         int[] digits = new int[10];
  5.         int[] cnt = new int[26];
  7.         for (char c : s.toCharArray()) {
  8.             cnt[c - 'a']++;
  9.         }
  11.         handleDigit(digits, cnt, 0, "zero", 'z');
  12.         handleDigit(digits, cnt, 2, "two", 'w');
  13.         handleDigit(digits, cnt, 4, "four", 'u');
  14.         handleDigit(digits, cnt, 6, "six", 'x');
  15.         handleDigit(digits, cnt, 8, "eight", 'g');
  16.         handleDigit(digits, cnt, 1, "one", 'o');
  17.         handleDigit(digits, cnt, 3, "three", 'h');
  18.         handleDigit(digits, cnt, 5, "five", 'f');
  19.         handleDigit(digits, cnt, 7, "seven", 's');
  20.         handleDigit(digits, cnt, 9, "nine", 'i');
  22.         for (int i = 0; i < 10; i++) {
  23.             for (int j = 0; j < digits[i]; j++) {
  24.                 sb.append(i);
  25.             }
  26.         }
  28.         return sb.toString();
  29.     }
  31.     private void handleDigit(int[] digits, int[] cnt,int digit, String word, char unique) {
  32.         if (cnt[unique - 'a'] > 0) {
  33.             int tmp = cnt[unique - 'a'];
  34.             digits[digit] += tmp;
  35.             for (char c : word.toCharArray()) {
  36.                 cnt[c - 'a'] -= tmp;
  37.             }
  38.         }
  39.     }
  40. }