[LeetCode] 423 Reconstruct Original Digits from English

Posted 2020-08-19 JavaNerd

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:

Input: "owoztneoer"

Output: "012"

 

Example 2:

Input: "fviefuro"

Output: "45"

---

解法：原来的字符串是乱序的，当遇到一个字母 e 的时候，并没有办法知道这个字符是属于 one、three、nine等中的一个。但是当遇到一个 z 的时候，肯定
是可以确定这个字符是属于 zero 的，因为z只出现在zero中，同理，x 也只出现在six中。
那么这就是一个优先级的问题，应该优先匹配那些属于少数数字的字符，例如优先寻找z，如果出现一个z，那么必须相应的出现e、r、o。如果字符串里面已经没有z了，
那么这个时候如果还出现了r，那么这个 r 就只可能属于 three、four中的一个了，依次类推，就能得到解法。

    for (char c : s.toCharArray()) {
            count[c]++;
        }
        int[] res = new int[10];
        char[] idx = {‘z‘, ‘x‘, ‘s‘, ‘v‘, ‘f‘, ‘r‘, ‘w‘, ‘g‘, ‘o‘, ‘i‘};
        //char[] idx = {‘z‘, ‘x‘, ‘s‘, ‘v‘, ‘f‘, ‘r‘, ‘w‘, ‘g‘, ‘o‘, ‘i‘};
        for (char i : idx) {
            while (count[i] > 0) {
                switch (i) {
                    case ‘z‘: //匹配z，减去zero
                        res[0]++;
                        count[‘z‘]--;
                        count[‘e‘]--;
                        count[‘r‘]--;
                        count[‘o‘]--;
                        break;   
                    case ‘x‘:   //匹配six,减去six
                        res[6]++;
                        count[‘s‘]--;
                        count[‘i‘]--;
                        count[‘x‘]--;
                        break;
                    case ‘s‘:  //匹配seven,因为之前优先匹配了six,如果这个时候,还有多余的s,那么肯定是seven
                        res[7]++;
                        count[‘s‘]--;
                        count[‘e‘]--;
                        count[‘v‘]--;
                        count[‘e‘]--;
                        count[‘n‘]--;
                        break;
                    case ‘v‘:   //匹配five,因为之前优先匹配了seven,如果这个时候还出现v,那么肯定是属于five
                        res[5]++;
                        count[‘f‘]--;
                        count[‘i‘]--;
                        count[‘v‘]--;
                        count[‘e‘]--;
                        break;
                    case ‘f‘:   //匹配four,因为之前匹配了four,那么这个时候还出现f,那么肯定属于four
                        res[4]++;
                        count[‘f‘]--;
                        count[‘o‘]--;
                        count[‘u‘]--;
                        count[‘r‘]--;
                        break;
                    case ‘r‘: //匹配three,因为之前匹配了four,还出现three,肯定属于three
                        res[3]++;
                        count[‘t‘]--;
                        count[‘h‘]--;
                        count[‘r‘]--;
                        count[‘e‘]--;
                        count[‘e‘]--;
                        break;
                    case ‘w‘:
                        res[2]++;
                        count[‘t‘]--;
                        count[‘w‘]--;
                        count[‘o‘]--;
                        break;
                    case ‘g‘:
                        res[8]++;
                        count[‘e‘]--;
                        count[‘i‘]--;
                        count[‘g‘]--;
                        count[‘h‘]--;
                        count[‘t‘]--;
                        break;
                    case ‘o‘:
                        res[1]++;
                        count[‘o‘]--;
                        count[‘n‘]--;
                        count[‘e‘]--;
                        break;
                    case ‘i‘:
                        res[9]++;
                        count[‘n‘]--;
                        count[‘i‘]--;
                        count[‘n‘]--;
                        count[‘e‘]--;
                        break;
                    default:
                        break;
                }
            }
        }
        //System.out.print(res[0]);
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i <= 9; i++) {
            for (int j = 0; j < res[i]; j++) {
                sb.append(i);
            }
        }
        return sb.toString();
    }