[Swift]LeetCode269. 外星人词典 $ Alien Dictionary

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There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"

Example 2:

Input:
[
  "z",
  "x"
]

Output: "zx"

Example 3:

""
""

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

---

有一种新的外来语言使用拉丁字母。 但是，您不知道字母之间的顺序。 您会从字典中收到一个非空单词列表，其中的单词按照新语言的规则按字典顺序排序。 导出这种语言的字母顺序。

例1：

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"

例2：

Input:
[
  "z",
  "x"
]

Output: "zx"

例3：

Input:
[
  "z",
  "x",
  "z"
] 

Output: "" 

说明：订单无效，因此请返回“”。
注意：

1、您可以假设所有字母都是小写的。
2、您可以假设如果a是b的前缀，则a必须出现在给定字典中的b之前。
3、如果订单无效，则返回空字符串。
4、可能有多个有效的字母顺序，返回其中任何一个都没问题。

---

Solution

class Solution {
    func alienOrder(_ words: [String]) -> String {
        var st:Set<[Character]> = Set<[Character]>()
        var ch:Set<Character> = Set<Character>()
        var ins:[Int] = [Int](repeating:0,count:256)
        var q:[Character] = [Character]()
        var res:String = String()
        for a in words
        {
            for c in a
            {
                ch.insert(c)
            }
        }
        for i in 0..<words.count - 1
        {
            let mn:Int = min(words[i].count, words[i + 1].count)
            var j:Int = 0
            while(j < min(words[i].count, words[i + 1].count))
            {
                if words[i][j] != words[i + 1][j]
                {
                    st.insert([words[i][j], words[i + 1][j]])
                    break
                }
                j += 1
            }
            if j == mn && words[i].count > words[i + 1].count
            {
                return String()
            }
        }
        for a in st
        {
            ins[a[1].ascii] += 1
        }
        for a in ch
        {
            if ins[a.ascii] == 0
            {
                q.append(a)
                res.append(a)
            }
        }
        while(!q.isEmpty)
        {
            let c:Character = q.removeFirst()
            for a in st
            {
                if a[0] == c
                {
                    ins[a[1].ascii] -= 1
                    if ins[a[1].ascii] == 0
                    {
                        q.append(a[1])
                        res.append(a[1])
                    }
                }
            }
        }
        return res.count == ch.count ? res : String()
    }
}

//String扩展
extension String {
    //subscript函数可以检索数组中的值
    //直接按照索引方式截取指定索引的字符
    subscript (_ i: Int) -> Character {
        //读取字符
        get {return self[index(startIndex, offsetBy: i)]}
    }
}

//Character扩展
extension Character
{
    //Character转ASCII整数值(定义小写为整数值)
    var ascii: Int {
        get {
            return Int(self.unicodeScalars.first?.value ?? 0)
        }
    }
}

---

点击：Playground测试

var sol = Solution()
print(sol.alienOrder(["wrt", "wrf", "er", "ett","rftt"]))
//Print wertf
print(sol.alienOrder(["z","x"]))
//Print zx
print(sol.alienOrder(["z","x","z"]))
//Print ""