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。
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 }
原文地址:https://www.cnblogs.com/strengthen/p/10090815.html
时间: 2024-07-31 21:28:06