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:

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"

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给定一个非空字符串，该字符串由0~9对应的单词组成，且各单词的字母顺序是打乱了的，要求按升序输出这个字符串所包含的数字。

注意：

　　1.所有字母均为小写；

　　2.保证输入的字符串由0~9对应的英文单词组成（乱序），即原字符串一定可以还原成数字表示且没有多余字符；

　　3.输入字符串长度小于50000。

 

分析：

　　先统计输入字符串中各字符的个数，分别计为c _a、c _b、……c _z，设0~9分别有n ₀、n ₁、……n ₉个，根据0~9对应英文单词（zero、one、two、three、four、five、six、seven、eight、nine）可知：

c _a=0

c _b=0

c _c=0

c _d=0

c _e=n ₀+n ₁+n ₃*2+n ₅+n ₇*2+n ₈+n ₉

c _f=n ₄+n ₅

c _g=n ₈

c _h=n ₃+n ₈

c _i=n ₅+n ₆+n ₈+n ₉

c _j=0

c _k=0

c _l=0

c _m=0

c _n=n ₁+n ₇+n ₉*2

c _o=n ₀+n ₁+n ₂+n ₄

c _p=0

c _q=0

c _r=n ₀+n ₃+n ₄

c _s=n ₆+n ₇

c _t=n ₂+n ₃+n ₈

c _u=n ₄

c _v=n ₅+n ₇

c _w=n ₂

c _x=n ₆

c _y=0

c _z=n0

 

由以上各等式可知：

n ₀=c _z

n ₂=c _w

n ₄=c _u

n ₆=c _x

n ₈=c _g

n ₅=c _f-n ₄

n ₁=c _o-n ₀-n ₂-n ₄

n ₃=c _h-n ₈

n ₇=c _s-n ₆

n ₉=c _i-n ₅-n ₆-n ₈

当然，求解等式并不只有这一种，比如由前面的分析可知，n ₇还可以由c _v-n ₅得出，选取任一种均可。

由以上分析编写相应代码如下：

string originalDigits(string s)
{
    string nums[10] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
    int counts[10] = { 0 };
    map<char, int> chars;
    for (char ch = 'a'; ch <= 'z'; ch++)
        chars[ch] = 0;
    for (char ch : s)
        chars[ch]++;
    counts[0] = chars['z'];
    counts[2] = chars['w'];
    counts[4] = chars['u'];
    counts[6] = chars['x'];
    counts[8] = chars['g'];
    counts[5] = chars['f'] - counts[4];
    counts[1] = chars['o'] - counts[0] - counts[2] - counts[4];
    counts[3] = chars['h'] - counts[8];
    counts[7] = chars['s'] - counts[6];
    counts[9] = chars['i'] - counts[5] - counts[6] - counts[8];
    string rst;
    for (int i = 0; i < 10; i++)
        if (counts[i] > 0)
            rst.append(counts[i], i + '0');
    return rst;
}