[Swift]LeetCode1160. 拼写单词 | Find Words That Can Be Formed by Characters
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You are given an array of strings words and a string chars.
A string is good if it can be formed by characters from chars (each character can only be used once).
Return the sum of lengths of all good strings in words.
Example 1:
Input: words = ["cat","bt","hat","tree"], chars = "atach"
Output: 6
Explanation:
The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
Example 2:
Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
Output: 10
Explanation:
The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
Note:
1 <= words.length <= 10001 <= words[i].length, chars.length <= 100- All strings contain lowercase English letters only.
给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
注意：每次拼写时，chars 中的每个字母都只能用一次。
返回词汇表 words 中你掌握的所有单词的 长度之和。
示例 1：
输入：words = ["cat","bt","hat","tree"], chars = "atach" 输出：6 解释： 可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。
示例 2：
输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr" 输出：10 解释： 可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。
提示：
1 <= words.length <= 10001 <= words[i].length, chars.length <= 100- 所有字符串中都仅包含小写英文字母
216ms
1 class Solution { 2 func countCharacters(_ words: [String], _ chars: String) -> Int { 3 var res = 0 4 var charMap = [Character: Int]() 5 for char in chars { 6 charMap[char] = charMap[char, default: 0] + 1 7 } 8 for word in words { 9 var shouldAdd = true 10 var charMapCopy = charMap 11 for char in word { 12 if let val = charMapCopy[char] { 13 if val > 0 { 14 charMapCopy[char] = val - 1 15 } else { 16 shouldAdd = false 17 break 18 } 19 } else { 20 shouldAdd = false 21 break 22 } 23 } 24 if shouldAdd { 25 res += word.count 26 } 27 } 28 return res 29 } 30 }
244ms
1 class Solution { 2 typealias Counter = [Character:Int] 3 var charsCounter: Counter = Counter() 4 var charsCount = 0 5 func countCharacters(_ words: [String], _ chars: String) -> Int { 6 var ans = 0 7 charsCounter = getWordCounter(chars) 8 charsCount = chars.count 9 for word in words { 10 if valid(word) { 11 print(word) 12 ans += word.count 13 } 14 } 15 return ans 16 } 17 18 private func getWordCounter(_ word: String) -> Counter { 19 var ans = Counter() 20 for ch in word { 21 if ans[ch] == nil { 22 ans[ch] = 1 23 } else { 24 ans[ch] = ans[ch]! + 1 25 } 26 } 27 return ans 28 } 29 private func valid(_ word: String) -> Bool { 30 if word.count > charsCount { 31 return false 32 } 33 var tempCharsCounter = charsCounter 34 for key in word { 35 if tempCharsCounter[key] == nil { 36 return false 37 } 38 tempCharsCounter[key] = tempCharsCounter[key]! - 1 39 if tempCharsCounter[key] == 0 { 40 tempCharsCounter.removeValue(forKey: key) 41 } 42 } 43 return true 44 } 45 }
288ms
1 class Solution { 2 func countCharacters(_ words: [String], _ chars: String) -> Int { 3 4 if words.count == 0 { return 0 } 5 if chars.count == 0 { return 0 } 6 7 var charDict = [Character: Int]() 8 var charStr = Array(chars) 9 10 for ch in charStr { 11 charDict[ch, default: 0] += 1 12 } 13 14 var result = 0 15 for word in words { 16 let wordChars = Array(word) 17 var wordDict = charDict 18 var canForm = true 19 20 for wch in wordChars { 21 guard var charFreq = wordDict[wch] else { 22 canForm = false 23 break 24 } 25 charFreq -= 1 26 27 if charFreq < 0 { 28 canForm = false 29 break 30 } 31 32 wordDict[wch] = charFreq 33 } 34 35 if canForm { 36 result += word.count 37 } 38 } 39 40 return result 41 } 42 }
292ms
1 class Solution { 2 func countCharacters(_ words: [String], _ chars: String) -> Int { 3 4 let chars1 = Array(chars) 5 var dic = [Character: Int]() 6 for c in chars1 { 7 dic[c, default: 0] += 1 8 } 9 10 var ans = 0 11 for word in words { 12 var tmpDic = dic 13 let chs = Array(word) 14 ans += chs.count 15 for c in chs { 16 if tmpDic[c ,default: 0] <= 0 { 17 ans -= chs.count 18 break 19 } 20 tmpDic[c ,default: 0] -= 1 21 } 22 } 23 return ans 24 } 25 }
520ms
1 class Solution { 2 func countCharacters(_ words: [String], _ chars: String) -> Int { 3 4 var count = 0 5 6 let mappedChars = chars.reduce(into: [:]) { 7 8 counts, character in 9 counts[character, default: 0] += 1 10 11 } 12 13 for word in words { 14 15 if isInChars(word, chars: mappedChars) { 16 count += word.count 17 } 18 19 } 20 21 return count 22 } 23 24 func isInChars(_ word: String, chars: [String.Element: Int]) -> Bool { 25 26 let mappedWord = word.reduce(into: [:]) { 27 counts, character in 28 counts[character, default: 0] += 1 29 } 30 31 for each in mappedWord { 32 33 if !chars.keys.contains(each.key) { 34 return false 35 } 36 37 if each.value > chars[each.key]! { 38 return false 39 } 40 41 } 42 43 return true 44 } 45 }
Runtime: 920 ms
Memory Usage: 21 MB
1 class Solution { 2 func countCharacters(_ words: [String], _ chars: String) -> Int { 3 var A:[Int:Int] = [Int:Int]() 4 let arrChars:[Int] = Array(chars).map{$0.ascii} 5 for c in arrChars 6 { 7 A[c - 97,default:0] += 1 8 } 9 var ret:Int = 0 10 for s in words 11 { 12 var cnt:[Int:Int] = [Int:Int]() 13 let arrS:[Int] = Array(s).map{$0.ascii} 14 for c in arrS 15 { 16 cnt[c - 97,default:0] += 1 17 } 18 var found:Bool = false 19 for k in 0..<26 20 { 21 if cnt[k,default:0] > A[k,default:0] 22 { 23 found = true 24 } 25 } 26 if !found 27 { 28 ret += s.count 29 } 30 } 31 return ret 32 } 33 } 34 35 //Character扩展 36 extension Character 37 { 38 //Character转ASCII整数值(定义小写为整数值) 39 var ascii: Int { 40 get { 41 return Int(self.unicodeScalars.first?.value ?? 0) 42 } 43 } 44 }
