[Swift]LeetCode58. 最后一个单词的长度 | Length of Last Word

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Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

Example:

Input: "Hello World"
Output: 5

---

给定一个仅包含大小写字母和空格 ' ' 的字符串，返回其最后一个单词的长度。

如果不存在最后一个单词，请返回 0 。

说明：一个单词是指由字母组成，但不包含任何空格的字符串。

示例:

输入: "Hello World"
输出: 5

---

8ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
    var count = 0
    for c in s.reversed() {
        if count == 0 && c == " " {  continue }
        if c == " " {
            return count
        }
        count += 1
    }
    return count
    }
}

---

12ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        
        var len = 0
        for str in s.reversed() {
            if str != " "{
                len += 1
            }
            if str == " " && len != 0 {
                break
            }
        }
        return len
    }
}

---

12ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        let s = Array(s)
        var tail = s.count - 1
        var result = 0
        
        while tail >= 0 && s[tail] == " " {
            tail -= 1
        }
        
        while tail >= 0 && s[tail] != " " {
            tail -= 1
            result += 1
        }
        
        return result
    }
}

---

16ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        if (s.isEmpty == false) {
            let fullNameArr = s.characters.split{$0 == " "}.map(String.init)
            if (fullNameArr.isEmpty) {
                return 0
            }
            return fullNameArr[fullNameArr.count - 1].count
        }
        return 0
    }
}

---

16ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        return s.split(separator: " ").last?.count ?? 0
    }
}

---

20ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        var result: Int = 0
        let array: Array = s.components(separatedBy: "\(" ")")
        for index in stride(from:array.count-1, to:-1, by:-1) {
            if array[index].count > 0 {
                result = array[index].count
                break
            }
        }
        return result
    }
}

---

 20ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
    var  sprlit  = s.components(separatedBy: .whitespacesAndNewlines)
    print(sprlit)
    guard s.count > 1 else{
        let inputWord = sprlit[0].trimmingCharacters(in: .whitespaces)
        return inputWord.count
    }
    var startCount = sprlit.count-1
    var lastWordCount = 0
    while   ( startCount >= 0){
         print("StartCount",startCount)
    var lastWord = sprlit[startCount]
  
    lastWord = lastWord.trimmingCharacters(in: .whitespacesAndNewlines)
        print("word:",lastWord)
     
            
        if(lastWord != ""){
            lastWordCount = lastWord.count
            return lastWordCount
         }
        startCount = startCount - 1
       lastWordCount = lastWord.count
    }
    return lastWordCount
 }
}

---

36ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        var lastWord = ""
        var restartLastWord = false
        for (index, c) in s.enumerated() {
            if c == " " {
                restartLastWord = true
            } else {
                if restartLastWord {
                    restartLastWord = false
                    lastWord = ""
                }
                lastWord += String(c)
            }
        }
        
        return lastWord.count
    }
}

---

36ms

class Solution {
    func lengthOfLastWord(_ s: String) -> Int {
        //将参数常量变为变量
        var word = s
        //最后一个单词的长度
        var len:Int = 0
        //字符串为空不存在最后一个单词返回 0 
        if  word.isEmpty{return 0}
        let wordCount:Int=word.count-1
        //倒序遍历
        for i in (0...wordCount).reversed()
        {
            //遍历到第一个空格退出循环
            if word[word.index(word.startIndex, offsetBy: i)] == " "
            {
                continue
            }
            //如果不是空格则继续遍历,i>=len 放前面, 否则 "q" 这样的字符串会越界
            while(i>=len && word[word.index(word.startIndex, offsetBy: i-len)] != " ")
            {
                len+=1 
            } 
           return len
        }
        return len
    }
}