Sometimes people repeat letters to represent extra feeling,such as "hello" -> "heeellooo","hi" -> "hiiii". Here,we have groups,of adjacent letters that are all the same character,and adjacent characters to the group are different. A group is extended if that group is length 3 or more,so "e" and "o" would be extended in the first example,and "i" would be extended in the second example. As another example,the groups of "abbcccaaaa" would be "a","bb","ccc",and "aaaa"; and "ccc" and "aaaa" are the extended groups of that string.
For some given string S,a query word is stretchy if it can be made to be equal to S by extending some groups. Formally,we are allowed to repeatedly choose a group (as defined above) of characters c,and add some number of the same character c to it so that the length of the group is 3 or more. Note that we cannot extend a group of size one like "h" to a group of size two like "hh" - all extensions must leave the group extended - ie.,at least 3 characters long.
Given a list of query words,return the number of words that are stretchy.
Example: Input: S = "heeellooo" words = ["hello","hi","helo"] Output: 1 Explanation: We can extend "e" and "o" in the word "hello" to get "heeellooo". We can‘t extend "helo" to get "heeellooo" because the group "ll" is not extended.
Notes:
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0 <= len(S) <= 100. -
0 <= len(words) <= 100. -
0 <= len(words[i]) <= 100. -
Sand all words inwordsconsist only of lowercase letters
有时候人们会用额外的字母来表示额外的情感，比如 "hello" -> "heeellooo","hi" -> "hiii"。我们将连续的相同的字母分组，并且相邻组的字母都不相同。我们将一个拥有三个或以上字母的组定义为扩张状态（extended），如第一个例子中的 "e" 和" o" 以及第二个例子中的 "i"。 此外，"abbcccaaaa" 将有分组 "a","dddd"；其中 "ccc" 和 "aaaa" 处于扩张状态。
对于一个给定的字符串 S ，如果另一个单词能够通过将一些字母组扩张从而使其和 S 相同，我们将这个单词定义为可扩张的（stretchy）。我们允许选择一个字母组（如包含字母 c ），然后往其中添加相同的字母 c 使其长度达到 3 或以上。注意，我们不能将一个只包含一个字母的字母组，如 "h"，扩张到一个包含两个字母的组，如 "hh"；所有的扩张必须使该字母组变成扩张状态（至少包含三个字母）。
输入一组单词，输出其中可扩张的单词数量。
示例： 输入： S = "heeellooo" words = ["hello","helo"] 输出：1 解释： 我们能通过扩张"hello"的"e"和"o"来得到"heeellooo"。 我们不能通过扩张"helo"来得到"heeellooo"因为"ll"不处于扩张状态。
说明：
-
0 <= len(S) <= 100。 -
0 <= len(words) <= 100。 -
0 <= len(words[i]) <= 100。 -
S和所有在words中的单词都只由小写字母组成。
1 class Solution { 2 func expressiveWords(_ S: String,_ words: [String]) -> Int { 3 let lenS = S.count 4 if lenS == 0 { // S 不能为空 5 return 0 6 } 7 var n = words.count,res = 0,arrS = Array(S) 8 for i in 0..<n { 9 let lenW = words[i].count 10 // words[i] 不能为空或者长度不小于S 11 if lenW == 0 || lenW >= lenS { 12 return 0 13 } 14 var j = 0,k = 0,streched = false,arrW = Array(words[i]) 15 while j < lenS && k < lenW { 16 var startS = j 17 var startW = k 18 if arrS[j] == arrW[k] { 19 while j < lenS - 1 && arrS[j] == arrS[j+1] { 20 j += 1 21 } 22 while k < lenW - 1 && arrW[k] == arrW[k+1] { 23 k += 1 24 } 25 // 如果S分组后的字符串是由某个字符重复2次及以上组成，则认为是可扩展 26 if j - startS >= 2 { 27 streched = true 28 } 29 //如果不可扩展，对应位置的字符串必须相同 30 else if j - startS != k - startW { 31 streched = false 32 break 33 } 34 } 35 else { 36 streched = false 37 break 38 } 39 j += 1 40 k += 1 41 } 42 if streched { 43 res += 1 44 } 45 } 46 return res 47 } 48 }
28ms
1 class Solution { 2 func expressiveWords(_ S: String,_ words: [String]) -> Int { 3 var sRLE = getRLE(Array(S)) 4 var result = 0 5 for word in words { 6 let wRLE = getRLE(Array(word)) 7 guard sRLE.count == wRLE.count else { 8 continue 9 } 10 var index = 0 11 while index < sRLE.count { 12 guard sRLE[index].0 == wRLE[index].0 else { 13 break 14 } 15 if !((sRLE[index].1 >= 3 && sRLE[index].1 >= wRLE[index].1) || sRLE[index].1 == wRLE[index].1) { 16 break 17 } 18 index += 1 19 } 20 if index == sRLE.count { 21 result += 1 22 } 23 } 24 return result 25 } 26 27 private func getRLE(_ chars: [Character]) -> [(Character,Int)] { 28 var start = 0 29 var pointer = 0 30 var result = [(Character,Int)]() 31 while pointer < chars.count { 32 while pointer < chars.count && chars[pointer] == chars[start] { 33 pointer += 1 34 } 35 result.append((chars[start],pointer - start)) 36 start = pointer 37 } 38 return result 39 } 40 }
40ms
1 class Solution { 2 func expressiveWords(_ S: String,_ words: [String]) -> Int { 3 func compress(_ s: String) -> [(Character,Int)] { 4 return s.reduce(into: [(Character,Int)]()) { 5 if let (char,count) = $0.last,char == $1 { 6 $0[$0.count-1] = (char,count+1) 7 } else { 8 $0.append(($1,1)) 9 } 10 } 11 } 12 13 func isExpressive(_ s: [(Character,Int)],_ t: [(Character,Int)]) -> Bool { 14 guard s.count == t.count else { return false } 15 for i in s.indices { 16 if s[i] == t[i] { continue } 17 if s[i].0 != t[i].0 || s[i].1 == 2 || s[i].1 < t[i].1 { return false } 18 } 19 return true 20 } 21 22 let set = Set(words) 23 let sCompressed = compress(S) 24 return words.filter{ isExpressive(sCompressed,compress($0)) }.count 25 } 26 }
1 class Solution { 2 func expressiveWords(_ S: String,_ words: [String]) -> Int { 3 var arrS:[Character] = Array(S) 4 var res:Int = 0 5 var m:Int = S.count 6 var n:Int = words.count 7 for word in words 8 { 9 var arrWord = Array(word) 10 var i:Int = 0 11 var j:Int = 0 12 while(i < m) 13 { 14 if j < word.count && arrS[i] == arrWord[j] 15 { 16 j += 1 17 } 18 else if i > 0 && arrS[i] == arrS[i - 1] && i + 1 < m && arrS[i] == arrS[i + 1] 19 { 20 i += 1 21 } 22 else if !(i > 1 && arrS[i] == arrS[i - 1] && arrS[i] == arrS[i - 2]) 23 { 24 break 25 } 26 i += 1 27 } 28 if i == m && j == word.count 29 { 30 res += 1 31 } 32 } 33 return res 34 } 35 }
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