摘要：如對(duì)應(yīng)的英文表達(dá)為并繼續(xù)亂序成。要求輸入亂序的英文表達(dá)式，找出其中包含的所有的數(shù)字，并按照從小到大輸出。思路和代碼首先將數(shù)字和英文表示列出來粗略一看，我們知道有許多字母只在一個(gè)英文數(shù)字中出現(xiàn)，比如只出現(xiàn)在中。

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(gè)非空的英文字符串，其中包含著亂序的阿拉伯?dāng)?shù)字的英文單詞。如012對(duì)應(yīng)的英文表達(dá)為zeroonetwo并繼續(xù)亂序成owoztneoer。要求輸入亂序的英文表達(dá)式，找出其中包含的所有0-9的數(shù)字，并按照從小到大輸出。

首先將數(shù)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(gè)英文數(shù)字中出現(xiàn)，比如z只出現(xiàn)在zero中。因此對(duì)于這種字母，它一旦出現(xiàn)，就意味著該數(shù)字一定出現(xiàn)了。
因此一輪過濾后可以得出只出現(xiàn)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(duì)剩下的數(shù)字字母過濾出只出現(xiàn)一次的字母：

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

最后對(duì)one和nine分別用o和i進(jìn)行區(qū)分即可。因此可以得出如下代碼：

    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
上面的代碼未免寫的太繁瑣了，對(duì)其進(jìn)一步優(yōu)化可以得到如下代碼：
    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();
    }