[Swift]LeetCode500. 键盘行 | Keyboard Row

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Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

Note:

1. You may use one character in the keyboard more than once.
2. You may assume the input string will only contain letters of alphabet.

---

给定一个单词列表，只返回可以使用在键盘同一行的字母打印出来的单词。键盘如下图所示。

 示例：

输入: ["Hello", "Alaska", "Dad", "Peace"]
输出: ["Alaska", "Dad"]

注意：

1. 你可以重复使用键盘上同一字符。
2. 你可以假设输入的字符串将只包含字母。

---

12ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        //char作key；行数作value
        var map:[Character:Int] = [Character:Int]()
        var res:[String] = [String]()
        //键盘三行
        var row1:String = "qwertyuiop"
        var row2:String  = "asdfghjkl"
        var row3:String = "zxcvbnm"
        makeMap(row1,1,&map)
        makeMap(row2,2,&map)
        makeMap(row3,3,&map)
        for word in words
        {
            //转换为小写
            var str = word.lowercased()
            //获取行数
            let rowNum:Int = map[str[str.startIndex]]!
            var checkAdd:Bool = true
            for char in str.characters
            {
                if rowNum != map[char]
                {
                    checkAdd = false
                    break
                }
            }
            if checkAdd
            {
                res.append(word)
            }         
        }
        return res       
    }
    //用于初始化字典
    func makeMap(_ str:String,_ num:Int,_ map :inout [Character:Int])
    {
        for char in str.characters
        {
            map[char] = num
        }
    }
}

---

8ms

class Solution {
    
var keyboardRows = ["qwertyuiopQWERTYUIOP","asdfghjklASDFGHJKL","zxcvbnmZXCVBNM"]

func findWords(_ words: [String]) -> [String] {

    var correctWords = [String]()
    
    var shouldAdd = true
    
    for word in words {
        let row = findRow(for: word)
        print("Trying: \(word)")
        shouldAdd = true
        for character in word {
            if !keyboardRows[row].contains(character) {
                print("Failed in: \(character) for row: \(row)")
                shouldAdd = false
            }
        }
        if shouldAdd {
            correctWords.append(word)
            print("CorrectWords: \(correctWords)")
        }
    }
    
    return correctWords
}

func findRow(for word: String) -> Int {
    for i in 0..<keyboardRows.count {
        if keyboardRows[i].contains(word.first!){
            return i
        }
    }
    return 0
}
}

---

12ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        guard words.count > 0 else { return [] }
        var dict: [Int : String] = [
            1:String("QWERTYUIOP".sorted()),
            2:String("ASDFGHJKL".sorted()),
            3:String("ZXCVBNM".sorted())
        ]
        var output: [String] = []
        for str in words {
            let processedStr = Set(str.uppercased()).sorted()
            guard processedStr.count > 0 else {
                output.append(str)
                continue
            }
            var key = 0
            if dict[1]!.contains(processedStr[0]) {
                key = 1
            } else if dict[2]!.contains(processedStr[0]) {
                key = 2
            } else if dict[3]!.contains(processedStr[0]) {
                key = 3
            }
            var valid = true
            for ch in processedStr {
                if !dict[key]!.contains(ch) {
                    valid = false
                }
            }
            if valid {
                output.append(str)
            }
        }
        return output
    }
}

---

12ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        let rowOne = "qwertyuiop"
        let rowTwo = "asdfghjkl"
        let rowThree = "zxcvbnm"

        return words.filter { contain(rowOne, $0) || contain(rowTwo, $0) || contain(rowThree, $0) }
    }

    private func contain(_ str: String, _ word: String) -> Bool {
        return word.lowercased().filter { !str.contains($0) }.count == 0
    }
}

---

16ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        var row1: Set<String> = ["Q", "W", "E", "R", "T", "Y", "U", "I", "O", "P"]
        var row2: Set<String> = ["A", "S", "D", "F", "G", "H", "J", "K", "L"]
        var row3: Set<String> = ["Z", "X", "C", "V", "B", "N", "M"]
        var ret: [String] = []
        
        for str in words {
            let char = str[str.index(str.startIndex, offsetBy: 0)]
            var sameRow = true
            var currentRow: Set<String>
            if row1.contains(String(char).capitalized) { currentRow = row1 }
            else if row2.contains(String(char).capitalized) { currentRow = row2 }
            else { currentRow = row3 }

            for char in str {                
                if !currentRow.contains(String(char).capitalized) { sameRow = false }
            }
            
            if sameRow { ret.append(str) }
        }
        
        return ret
    }
}

---

20ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        let line1 : String = "qwertyuiop"
        let line2 : String = "asdfghjkl"
        let line3 : String = "zxcvbnm"
        var results : [String] = []
        
        var containLine : Int = 0
        for word in words {
            containLine = 0
            for w in word {
                if line1.contains(w) {
                    if containLine == 0 {
                        containLine = 1
                    }else if containLine != 1{
                        containLine = -1
                        break
                    }
                    continue
                }
                if line2.contains(w) {
                    if containLine == 0 {
                        containLine = 2
                    }else if containLine != 2{
                        containLine = -1
                        break
                    }
                    continue
                }
                if line3.contains(w) {
                    if containLine == 0 {
                        containLine = 3
                    }else if containLine != 3{
                        containLine = -1
                        break
                    }
                    continue
                }
            }
            if containLine != -1 {
                results.append(word)
            }
        }
        return results
    }
}

---

28ms

class Solution {
    func findWords(_ words: [String]) -> [String] {
        var res = [String]()
        let dic = ["q":0,"w":0,"e":0,"r":0,"t":0,"y":0,"u":0,"i":0,"o":0,"p":0,
                   "a":1,"s":1,"d":1,"f":1,"g":1,"h":1,"j":1,"k":1,"l":1,
                   "z":2,"x":2,"c":2,"v":2,"b":2,"n":2,"m":2,]
        for s in words{
            let low = s.lowercased()
            var line:Int?
            var bool = true
            for c in low{
                let num = dic[String(c)]
                if line == nil{
                    line = num
                }else if line != num{
                    bool = false
                    break
                }
            }
            if bool {
                res.append(s)
            }
        }
        return res
    }
}