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Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.git
Example 1:github
Input: s = "abpcplea", d = ["ale","apple","monkey","plea"] Output: "apple"
Example 2:微信
Input: s = "abpcplea", d = ["a","b","c"] Output: "a"
Note:app
給定一個字符串和一個字符串字典,找到字典裏面最長的字符串,該字符串能夠經過刪除給定字符串的某些字符來獲得。若是答案不止一個,返回長度最長且字典順序最小的字符串。若是答案不存在,則返回空字符串。spa
示例 1:code
輸入: s = "abpcplea", d = ["ale","apple","monkey","plea"] 輸出: "apple"
示例 2:orm
輸入: s = "abpcplea", d = ["a","b","c"] 輸出: "a"
說明:htm
340msblog
1 class Solution { 2 func findLongestWord(_ s: String, _ d: [String]) -> String { 3 var result : [Character] = [] 4 let s = Array(s) 5 for str in d { 6 var str = Array(str) 7 if str.count < result.count{ 8 continue 9 } 10 11 if canBeFoundFrom(s: s, with: Array(str)){ 12 if str.count > result.count{ 13 result = str 14 }else if str.count == result.count{ 15 result = lexicoOrder(result, str) 16 } 17 18 } 19 } 20 21 return String(result) 22 } 23 24 func lexicoOrder(_ first : [Character], _ second : [Character])->[Character]{ 25 26 var firstIndex : Int = 0 27 var secondIndex : Int = 0 28 29 while firstIndex < first.count{ 30 if first[firstIndex] == second[secondIndex]{ 31 firstIndex += 1 32 secondIndex += 1 33 }else if first[firstIndex] < second[secondIndex]{ 34 return first 35 }else{ 36 return second 37 } 38 } 39 return first 40 } 41 42 func canBeFoundFrom(s : [Character], with d : [Character])->Bool{ 43 44 45 var sIndex : Int = 0 46 var dIndex : Int = 0 47 while sIndex < s.count{ 48 if d[dIndex] == s[sIndex]{ 49 dIndex += 1 50 } 51 sIndex += 1 52 if dIndex == d.count{ 53 return true 54 } 55 } 56 57 return dIndex == d.count 58 } 59 }
532ms
1 class Solution { 2 func findLongestWord(_ s: String, _ d: [String]) -> String { 3 if d.count == 0 { 4 return "" 5 } 6 7 var long = "" 8 for items in d { 9 let i = long.count 10 let j = items.count 11 if i > j || (i == j && long < items) { 12 continue 13 } 14 if isValid(s: s, d: items) { 15 long = items 16 } 17 } 18 19 return long 20 } 21 22 fileprivate func isValid(s: String, d: String) -> Bool { 23 var i = 0 24 var j = 0 25 var source = Array(s) 26 var target = Array(d) 27 var result = "" 28 while i < source.count && j < target.count { 29 if source[i] == target[j] { 30 j += 1 31 result.append(source[i]) 32 } 33 i += 1 34 } 35 return j == target.count 36 } 37 }
840ms
1 class Solution { 2 func findLongestWord(_ s: String, _ d: [String]) -> String { 3 4 let sortedD = d.sorted { (first, second) -> Bool in 5 if first.count > second.count{ 6 return true 7 }else if first.count == second.count{ 8 return first < second 9 }else{ 10 return false 11 } 12 } 13 14 for str in sortedD{ 15 if possibleToForm(s, str: str){ 16 return str 17 } 18 } 19 return "" 20 } 21 22 func possibleToForm(_ base : String, str : String)->Bool{ 23 guard base.count >= str.count else { 24 return false 25 } 26 27 var baseIndex : String.Index = base.startIndex 28 var strIndex : String.Index = str.startIndex 29 30 while baseIndex != base.endIndex && strIndex != str.endIndex{ 31 if str[strIndex] == base[baseIndex]{ 32 strIndex = str.index(after: strIndex) 33 } 34 baseIndex = base.index(after: baseIndex) 35 } 36 37 return strIndex == str.endIndex 38 } 39 }
1 class Solution { 2 func findLongestWord(_ s: String, _ d: [String]) -> String { 3 var res:String = String() 4 for str in d 5 { 6 var arr:[Character] = Array(str) 7 var i:Int = 0 8 for c in s.characters 9 { 10 if i < str.count && c == arr[i] 11 { 12 i += 1 13 } 14 } 15 if i == str.count && str.count >= res.count 16 { 17 if str.count > res.count || str < res 18 { 19 res = str 20 } 21 } 22 } 23 return res 24 } 25 }