[Swift]LeetCode290. 单词模式 | Word Pattern

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Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Example 1:

"abba"
"dog cat cat dog"

Example 2:

"abba"
"dog cat cat fish"

Example 3:

"aaaa"
"dog cat cat dog"

Example 4:

"abba"
"dog dog dog dog"

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

---

 给定一种 pattern(模式) 和一个字符串 str ，判断 str 是否遵循相同的模式。

这里的遵循指完全匹配，例如， pattern 里的每个字母和字符串 str 中的每个非空单词之间存在着双向连接的对应模式。

示例1:

"abba"
"dog cat cat dog"

示例 2:

"abba"
"dog cat cat fish"

示例 3:

"aaaa"
"dog cat cat dog"

示例 4:

"abba"
"dog dog dog dog"

说明:
你可以假设 pattern 只包含小写字母， str 包含了由单个空格分隔的小写字母。    

---

 12ms

class Solution {
    func wordPattern(_ pattern: String, _ str: String) -> Bool {
        var strArr = str.components(separatedBy: " ");
        if pattern.count != strArr.count {
            return false
        }
        var keyValueDic = Dictionary<Character, String>()
        var keyValueDic2 = Dictionary<String, Character>()
        for (index,key) in pattern.enumerated() {
            let tempStr = keyValueDic[key]
            if tempStr == nil {
                if keyValueDic2[strArr[index]] != nil {
                    return false;
                }
                keyValueDic[key] = strArr[index]
                keyValueDic2[strArr[index]] = key
            }
            else {
                if tempStr == strArr[index] && keyValueDic2[tempStr!] == key {
                    continue
                }
                else {
                    return false
                }
            }
        }
        return true
    }
}

---

12ms

class Solution {
 func wordPattern(_ pattern: String, _ str: String) -> Bool {
    
    let patternArr = Array(pattern)
    let strArr = str.split(separator: " ").map{String($0)}
    
    var patternDict = [Character:Int]()
    var strDict = [String:Int]()
    
    var tempP = 0
    var tempS = 0

    for i in patternArr {
        if patternDict[i] == nil {
            patternDict[i] = tempP
            tempP += 1
        }
    }

    for i in strArr {
        if strDict[i] == nil {
            strDict[i] = tempS
            tempS += 1
        }
    }
    
    var resultPattern = [Int]()
    var resultStr = [Int]()
    
    for i in patternArr {
        resultPattern.append(patternDict[i]!)
    }
    
    for i in strArr {
        resultStr.append(strDict[i]!)
    }
    
    print(resultPattern)
    print(resultStr)
    print(resultPattern == resultStr)
    return resultPattern == resultStr
  }
}

---

16ms

class Solution {
    func wordPattern(_ pattern: String, _ str: String) -> Bool {
        let wordArr = str.components(separatedBy: " ")
        if pattern.count != wordArr.count {
            return false
        }

        var patternMap: [String: String] = [String: String]()

        for (n,x) in pattern.enumerated() {
            let strX = String(x)
            if patternMap[strX] == nil {//没有strx这个key
                let word = wordArr[n]
                let flag = patternMap.values.contains(word)
                if !flag {
                    patternMap[strX] = word
                }else {
                    return false
                }
            }else {
                if patternMap[strX] != wordArr[n] {
                    return false
                }
            }
        }
        return true
    }
}

---

20ms

class Solution {
    func wordPattern(_ pattern: String, _ str: String) -> Bool {
        let strArr = str.split(separator: " ")
        var dic = [Character: String]()
        if strArr.count != pattern.characters.count || Set(strArr).count != Set(pattern.characters).count{
        return false
    }
    
    for (i, v) in pattern.characters.enumerated(){
        if dic[v] == nil{
            dic[v] = String(strArr[i])
        }else{
            if dic[v] != String(strArr[i]){
                return false
            }
        }
    }
    return true
    }
}