leetcode423. Reconstruct Original Digits from English

摘要： 題目要求 Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order. Note: Input...

題目要求

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"

一個非空的英文字串，其中包含著亂序的阿拉伯數字的英文單詞。如012 對應的英文表達為zeroonetwo 並繼續亂序成owoztneoer 。要求輸入亂序的英文表達式，找出其中包含的所有0-9 的數字，並按照從小到大輸出。

思路和程式碼

首先將數字和英文表示列出來：

0 zero
1 one
2 two
3 three
4 four
5 five
6 six
7 seven
8 eight
9 nine

粗略一看，我們知道有許多字母只在一個英文數字中出現，比如z只出現在zero中。因此對於這種字母，它一旦出現，就意味著該數字一定出現了。

因此一輪過濾後可以得出只出現一次的字母如下：

0 zero -> z
1 one
2 two -> w
3 three
4 four -> u
5 five
6 six -> x
7 seven
8 eight
9 nine

再對剩下的數字字母過濾出只出現一次的字母：

1 one 
3 three -> r
5 five -> f
7 seven -> s
8 eight -> g
9 nine

最後對one和nine分別用o和i進行區分即可。因此可以得出如下程式碼：

public String originalDigits(String s) {
int[] letterCount = new int[26];
for(char c : s.toCharArray()) {
letterCount[c-'a']++;
}

int[] result = new int[10];

//zero
if((result[2] = letterCount['z'-'a']) != 0) {
result[0] = letterCount['z' - 'a'];
letterCount['z'-'a'] = 0;
letterCount['e'-'a'] -= result[0];
letterCount['r'-'a'] -= result[0];
letterCount['o'-'a'] -= result[0];
}
//two
if((result[2] = letterCount['w'-'a']) != 0) {
letterCount['t'-'a'] -= result[2];
letterCount['w'-'a'] = 0;
letterCount['o'-'a'] -= result[2];
}
//four
if((result[4] = letterCount['u'-'a']) != 0) {
letterCount['f'-'a'] -= result[4];
letterCount['o'-'a'] -= result[4];
letterCount['u'-'a'] -= result[4];
letterCount['r'-'a'] -= result[4];
}
//five
if((result[5] = letterCount['f'-'a']) != 0) {
letterCount['f'-'a'] -= result[5];
letterCount['i'-'a'] -= result[5];
letterCount['v'-'a'] -= result[5];
letterCount['e'-'a'] -= result[5];
}
//six
if((result[6] = letterCount['x'-'a']) != 0) {
letterCount['s'-'a'] -= result[6];
letterCount['i'-'a'] -= result[6];
letterCount['x'-'a'] -= result[6];
}
//seven
if((result[7] = letterCount['s'-'a']) != 0) {
letterCount['s'-'a'] -= result[7];
letterCount['e'-'a'] -= result[7] * 2;
letterCount['v'-'a'] -= result[7];
letterCount['n'-'a'] -= result[7];
}
//one
if((result[1] = letterCount['o'-'a']) != 0) {
letterCount['o'-'a'] -= result[1];
letterCount['n'-'a'] -= result[1];
letterCount['e'-'a'] -= result[1];
}
//eight
if((result[8] = letterCount['g'-'a']) != 0) {
letterCount['e'-'a'] -= result[8];
letterCount['i'-'a'] -= result[8];
letterCount['g'-'a'] -= result[8];
letterCount['h'-'a'] -= result[8];
letterCount['t'-'a'] -= result[8];
}
//nine
if((result[9] = letterCount['i'-'a']) != 0) {
letterCount['n'-'a'] -= result[9] * 2;
letterCount['i'-'a'] -= result[9];
letterCount['e'-'a'] -= result[9];
}
result[3] = letterCount['t'-'a'];
StringBuilder sb = new StringBuilder();
for(int i = 0 ; i<result.length ; i++) {
for(int j = 0 ; j<result[i] ; j++) {
sb.append(i);
}
}
return sb.toString();
}

上面的程式碼未免寫的太繁瑣了，對其進一步優化可以得到如下程式碼：

public String originalDigits2(String s) {
int[] alphabets = new int[26];
for (char ch : s.toCharArray()) {
alphabets[ch - 'a'] += 1;
}

int[] digits = new int[10];

digits[0] = alphabets['z' - 'a'];
digits[2] = alphabets['w' - 'a'];
digits[6] = alphabets['x' - 'a'];
digits[8] = alphabets['g' - 'a'];
digits[7] = alphabets['s' - 'a'] - digits[6];
digits[5] = alphabets['v' - 'a'] - digits[7];
digits[3] = alphabets['h' - 'a'] - digits[8];
digits[4] = alphabets['f' - 'a'] - digits[5];
digits[9] = alphabets['i' - 'a'] - digits[6] - digits[8] - digits[5];
digits[1] = alphabets['o' - 'a'] - digits[0] - digits[2] - digits[4];

StringBuilder sb = new StringBuilder();
for (int d = 0; d < 10; d++) {
for (int count = 0; count < digits[d]; count++) sb.append(d);
}

return sb.toString();
}