Reconstruct Original Digits from English

udpate Aug 6,2017 16:42

LeetCode

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"

Basic Idea:

The even digits all have a unique letter while the odd digits all don't:

zero: Only digit with z
two: Only digit with w
four: Only digit with u
six: Only digit with x
eight: Only digit with g

The odd ones for easy looking, each one's letters all also appear in other digit words: one, three, five, seven, nine

Java Code (来自leetcode discussion):

    public class Solution {
        public String originalDigits(String s) {
            int[] count = new int[10];
            for (int i = 0; i < s.length(); i++){
                char c = s.charAt(i);
                if (c == 'z') count[0]++;
                if (c == 'w') count[2]++;
                if (c == 'x') count[6]++;
                if (c == 's') count[7]++; //7-6
                if (c == 'g') count[8]++;
                if (c == 'u') count[4]++; 
                if (c == 'f') count[5]++; //5-4
                if (c == 'h') count[3]++; //3-8
                if (c == 'i') count[9]++; //9-8-5-6
                if (c == 'o') count[1]++; //1-0-2-4
            }
            count[7] -= count[6];
            count[5] -= count[4];
            count[3] -= count[8];
            count[9] = count[9] - count[8] - count[5] - count[6];
            count[1] = count[1] - count[0] - count[2] - count[4];
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i <= 9; i++){
                for (int j = 0; j < count[i]; j++){
                    sb.append(i);
                }
            }
            return sb.toString();
        }
    }