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  • [Swift]LeetCode1153. 字符串转化 | String Transforms Into Another String

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    Given two strings str1 and str2 of the same length, determine whether you can transform str1 into str2 by doing zero or more conversions.

    In one conversion you can convert all occurrences of one character in str1 to any other lowercase English character.

    Return true if and only if you can transform str1 into str2.

    Example 1:

    Input: str1 = "aabcc", str2 = "ccdee"
    Output: true
    Explanation: Convert 'c' to 'e' then 'b' to 'd' then 'a' to 'c'. Note that the order of conversions matter.
    

    Example 2:

    Input: str1 = "leetcode", str2 = "codeleet"
    Output: false
    Explanation: There is no way to transform str1 to str2.

    Note:

    1. 1 <= str1.length == str2.length <= 10^4
    2. Both str1 and str2 contain only lowercase English letters.

    给出两个长度相同的字符串,分别是 str1 和 str2。请你帮忙判断字符串 str1 能不能在 零次 或 多次 转化后变成字符串 str2

    每一次转化时,将会一次性将 str1 中出现的 所有 相同字母变成其他 任何 小写英文字母(见示例)。

    只有在字符串 str1 能够通过上述方式顺利转化为字符串 str2 时才能返回 True,否则返回 False。​​

    示例 1:

    输入:str1 = "aabcc", str2 = "ccdee"
    输出:true
    解释:将 'c' 变成 'e',然后把 'b' 变成 'd',接着再把 'a' 变成 'c'。注意,转化的顺序也很重要。
    

    示例 2:

    输入:str1 = "leetcode", str2 = "codeleet"
    输出:false
    解释:我们没有办法能够把 str1 转化为 str2。

    提示:

    1. 1 <= str1.length == str2.length <= 10^4
    2. str1 和 str2 中都只会出现 小写英文字母

    72ms

     1 class Solution {
     2     func canConvert(_ str1: String, _ str2: String) -> Bool {
     3         if str1 == str2 {return true}
     4         let n:Int = str1.count
     5         var arr:[Int] = [Int](repeating:-1,count:26)
     6         let arrS:[Int] = Array(str1).map{$0.ascii}
     7         let arrT:[Int] = Array(str2).map{$0.ascii}
     8         for i in 0..<n
     9         {
    10             var x:Int = arrS[i] - 97
    11             var y:Int = arrT[i] - 97
    12             if arr[x] == -1
    13             {
    14                 arr[x] = y
    15             }
    16             else if arr[x] != y
    17             {
    18                 return false
    19             }
    20         }
    21         var has:Int = 0
    22         for i in 0..<26
    23         {
    24             if arr[i] != -1
    25             {
    26                 has += 1
    27             }            
    28         }
    29         var flag:Int = 0
    30         for i in 0..<26
    31         {
    32             for j in (i + 1)..<26
    33             {
    34                 if arr[i] != -1 && arr[j] != -1 && arr[i] == arr[j]
    35                 {
    36                     flag = 1
    37                 }
    38             }
    39         }
    40         if has != 26 || flag != 0 {return true}
    41         return false
    42     }
    43 }
    44 
    45 //Character扩展 
    46 extension Character  
    47 {  
    48   //Character转ASCII整数值(定义小写为整数值)
    49    var ascii: Int {
    50        get {
    51            return Int(self.unicodeScalars.first?.value ?? 0)
    52        }       
    53     }
    54 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/11333863.html
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