[Swift]LeetCode953. 验证外星语词典 | Verifying an Alien Dictionary
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In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order
. The order
of the alphabet is some permutation of lowercase letters.
Given a sequence of words
written in the alien language, and the order
of the alphabet, return true
if and only if the given words
are sorted lexicographicaly in this alien language.
Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
Note:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
- All characters in
words[i]
andorder
are english lowercase letters.
某种外星语也使用英文小写字母,但可能顺序 order
不同。字母表的顺序(order
)是一些小写字母的排列。
给定一组用外星语书写的单词 words
,以及其字母表的顺序 order
,只有当给定的单词在这种外星语中按字典序排列时,返回 true
;否则,返回 false
。
示例 1:
输入:words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz" 输出:true 解释:在该语言的字母表中,'h' 位于 'l' 之前,所以单词序列是按字典序排列的。
示例 2:
输入:words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz" 输出:false 解释:在该语言的字母表中,'d' 位于 'l' 之后,那么 words[0] > words[1],因此单词序列不是按字典序排列的。
示例 3:
输入:words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz" 输出:false 解释:当前三个字符 "app" 匹配时,第二个字符串相对短一些,然后根据词典编纂规则 "apple" > "app",因为 'l' > '∅',其中 '∅' 是空白字符,定义为比任何其他字符都小(更多信息)。
提示:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
- 在
words[i]
和order
中的所有字符都是英文小写字母。
28ms
1 class Solution { 2 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 3 var ret = true 4 5 for i in 1 ..< words.count { 6 ret = isInOrder(words[i-1], words[i], order) 7 if !ret { return false } 8 } 9 10 return true 11 } 12 13 func isInOrder(_ word1: String, _ word2: String, _ order: String) -> Bool{ 14 15 let word1 = Array(word1) 16 let word2 = Array(word2) 17 18 var i = 0, j = 0 19 while i < word1.count && j < word2.count { 20 if order.firstIndex(of: word1[i])! == order.firstIndex(of: word2[j])! { 21 i += 1 22 j += 1 23 }else if order.firstIndex(of: word1[i])! < order.firstIndex(of: word2[j])!{ 24 return true 25 }else { 26 return false 27 } 28 } 29 30 if word1.count == word2.count { 31 return true 32 } else if word1.count > word2.count{ 33 return false 34 } 35 36 return true 37 } 38 }
32ms
1 class Solution { 2 private var map = [Character: Int]() 3 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 4 for (index, char) in order.enumerated() { 5 map[char] = index 6 } 7 8 for (index, word) in words.enumerated() where index < words.count-1 { 9 if !wordsInOrder(word1: word, word2: words[index+1]) { 10 return false 11 } 12 } 13 14 return true 15 } 16 17 private func wordsInOrder(word1: String, word2: String) -> Bool { 18 let one = Array(word1) 19 let two = Array(word2) 20 21 var index = 0 22 var areEqual = true 23 24 while index < one.count || index < two.count { 25 if areEqual && two.count == index && one.count > index { 26 return false 27 } else if areEqual && one.count == index && two.count > index { 28 return true 29 } 30 31 let first = one[index] 32 let second = two[index] 33 34 index += 1 35 36 if first == second { 37 continue 38 } else { 39 areEqual = false 40 } 41 42 if map[first]! < map[second]! { 43 return true 44 } else if map[first]! == map[second]! { 45 continue 46 } else { 47 return false 48 } 49 } 50 51 return true 52 } 53 }
40ms
1 class Solution { 2 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 3 var dict: [Character: Int] = [:] 4 5 for (i, char) in order.enumerated() { 6 dict[char] = i 7 } 8 9 outer: for i in 0..<words.count - 1 { 10 var j = 0 11 let currWord = Array(words[i]) 12 let nextWord = Array(words[i + 1]) 13 14 while j < currWord.count && j < nextWord.count { 15 let currVal = dict[currWord[j]]! 16 let nextVal = dict[nextWord[j]]! 17 if currVal > nextVal { 18 return false 19 } else if currVal < nextVal { 20 continue outer 21 } 22 j += 1 23 } 24 25 return false 26 } 27 28 return true 29 } 30 }
44ms
1 class Solution { 2 3 var orderDic : [Character:Int] = [:] 4 5 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 6 var currentOrder = 0 7 for (idx,ord) in order.enumerated() { 8 orderDic[ord] = idx 9 } 10 11 for i in 0..<words.count-1 { 12 if !self.isOrdered(words[i], words[i+1]) { 13 return false 14 } 15 } 16 17 return true 18 } 19 20 func isOrdered(_ word1 : String, _ word2: String) -> Bool { 21 let word2Chars = Array(word2) 22 for (idx, char) in word1.enumerated() { 23 if idx >= word2.count { 24 return false 25 } 26 27 if orderDic[char]! > orderDic[word2Chars[idx]]! { 28 return false 29 } else if orderDic[char]! < orderDic[word2Chars[idx]]! { 30 return true 31 } 32 } 33 34 return true 35 } 36 }
48ms
1 class Solution { 2 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 3 4 var map = [Character: Int]() 5 let orderChars = Array(order) 6 for i in 0..<orderChars.count { 7 map[orderChars[i]] = i 8 } 9 10 for i in 1..<words.count { 11 if !isAscend(words[i-1], words[i], map: map) { 12 return false 13 } 14 } 15 return true 16 } 17 18 func isAscend(_ str1: String, _ str2: String, map:[Character: Int]) -> Bool { 19 20 let char1 = Array(str1) 21 let char2 = Array(str2) 22 23 let len = min(char1.count, char2.count) 24 25 for i in 0..<len { 26 if map[char1[i]]! > map[char2[i]]!{ 27 return false 28 } 29 if map[char1[i]]! < map[char2[i]]!{ 30 return true 31 } 32 } 33 return char1.count <= char2.count 34 } 35 }
52ms
1 class Solution { 2 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 3 var index = 0 4 var dict = [Character: Int]() 5 for char in order { 6 dict[char] = index 7 index += 1 8 } 9 // 10 if words.count == 1 {return true} 11 var base = words[0] 12 for i in 1..<words.count { 13 var curIndex = words[i].startIndex 14 while !base.isEmpty && base != words[i] { 15 16 if dict[base.first!]! < dict[words[i][curIndex]]! { 17 base = words[i] 18 continue 19 } else if dict[base.first!] == dict[words[i].first!] { 20 base.removeFirst() 21 if curIndex == words[i].index(words[i].startIndex, offsetBy: words[i].count - 1) && !base.isEmpty { return false } 22 curIndex = words[i].index(after: curIndex) 23 24 } else { 25 return false 26 } 27 28 } 29 } 30 return true 31 } 32 }
72ms
1 class Solution { 2 func isAlienSorted(_ words: [String], _ order: String) -> Bool { 3 var words = words 4 var a:[Int] = [Int](repeating:0,count:256) 5 var i:Int = 0 6 var j:Int = 0 7 for i in 0..<order.count 8 { 9 a[order[i].ascii] = i 10 } 11 for i in 0..<words.count 12 { 13 for j in 0..<words[i].count 14 { 15 //a:97 16 words[i][j] = (a[words[i][j].ascii] + 97).ASCII 17 } 18 } 19 for i in 0..<(words.count - 1) 20 { 21 if words[i] > words[i+1] 22 { 23 return false 24 } 25 } 26 return true 27 } 28 } 29 extension String { 30 //subscript函数可以检索数组中的值 31 //直接按照索引方式截取指定索引的字符 32 subscript (_ i: Int) -> Character { 33 //读取字符 34 get {return self[index(startIndex, offsetBy: i)]} 35 36 //修改字符 37 set 38 { 39 var str:String = self 40 var index = str.index(startIndex, offsetBy: i) 41 str.remove(at: index) 42 str.insert(newValue, at: index) 43 self = str 44 } 45 } 46 } 47 48 //Character扩展方法 49 extension Character 50 { 51 //属性:ASCII整数值(定义小写为整数值) 52 var ascii: Int { 53 get { 54 let s = String(self).unicodeScalars 55 return Int(s[s.startIndex].value) 56 } 57 } 58 } 59 60 //Int扩展方法 61 extension Int 62 { 63 //属性:ASCII值(定义大写为字符值) 64 var ASCII:Character 65 { 66 get {return Character(UnicodeScalar(self)!)} 67 } 68 }