423. Reconstruct Original Digits from English
Given a non-empty string containing an out-of-order English representation of digits
0-9
, output the digits in ascending order.Note:
- Input contains only lowercase English letters.
- 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.
- Input length is less than 50,000.
Example 1:
Input: "owoztneoer" Output: "012"
Example 2:
Input: "fviefuro" Output: "45"
Approach #1: Math. [Java]
class Solution { public String originalDigits(String s) { int[] count = new int[10]; for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) == 'z') count[0]++; if (s.charAt(i) == 'w') count[2]++; if (s.charAt(i) == 'u') count[4]++; if (s.charAt(i) == 'x') count[6]++; if (s.charAt(i) == 'g') count[8]++; if (s.charAt(i) == 's') count[7]++; // 7 - 6 if (s.charAt(i) == 'f') count[5]++; // 5 - 4 if (s.charAt(i) == 'i') count[9]++; // 9 - 8 - 6 - 5 if (s.charAt(i) == 'o') count[1]++; // 1 - 0 - 2 - 4 if (s.charAt(i) == 'h') count[3]++; // 3 - 8 } count[7] -= count[6]; count[5] -= count[4]; count[3] -= count[8]; count[9] = count[9] - count[8] - count[6] - count[5]; count[1] = count[1] - count[0] - count[2] - count[4]; System.out.println(count[1]); StringBuilder sb = new StringBuilder(); for (int i = 0; i < 10; ++i) { for (int j = 0; j < count[i]; ++j) { sb.append(i); } } return sb.toString(); } }
Analysis:
zero, one, two, three, four, five, six, seven, eight, nine
We can find character 'z' only appear in zero, character 'w' only appear in two .....
So we can count the special character to find how many times the number appeared.
If there don't have special character in the number, we can calculate by subtracting the repeated character.
Reference:
永远渴望,大智若愚(stay hungry, stay foolish)