[Swift]LeetCode387. 字符串中的第一个唯一字符 | First Unique Character in a String

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Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

 

Note: You may assume the string contain only lowercase letters.

给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。

案例:

s = "leetcode"
返回 0.

s = "loveleetcode",
返回 2.

 

注意事项：您可以假定该字符串只包含小写字母。

---

class Solution {
    func firstUniqChar(_ s: String) -> Int {
       //把英文字符都存入数组中
        var arr = Array<Int>(repeating: 0, count: 26)
        for asc in s.unicodeScalars
        {
            let num:Int = Int(asc.value - 97)
            //记录字符出现的次数
            arr[num] += 1
        }
        //再次循环字符串，使用enumerated()获取到字符串的索引
        for (index, asc) in s.unicodeScalars.enumerated()
        {
            let count = arr[Int(asc.value - 97)]
            if count == 1 {
                return index
            }
        }
        return -1
    }
}

---

104ms

class Solution {
    func firstUniqChar(_ s: String) -> Int {
        var list:[Int] = [Int](repeating: 0, count: 26)
        for charCode in s.unicodeScalars {
            let tempIndex:Int = Int(charCode.value) - 97
            list[tempIndex] += 1
        }
        
        var index = 0
        for charCode in s.unicodeScalars {
            let tempIndex:Int = Int(charCode.value) - 97
            if list[tempIndex] == 1 {
                return index
            }
            index += 1
        }
        return -1
    }
}

---

128ms

class Solution {
  func firstUniqChar(_ s: String) -> Int {
      var countFor = Array(repeating:0, count:26)
      let offset = 97
      for charCode in s.utf8 {
          let adjustedCharCode = Int(charCode) - offset
          countFor[adjustedCharCode] += 1
      }
      for (i, charCode) in s.utf8.enumerated() {
          let adjustedCharCode = Int(charCode) - offset
          if countFor[adjustedCharCode] == 1 {
              return i
          }
      }
      return -1
  }
}