[Swift]LeetCode459. 重複的子字符串 | Repeated Substring Pattern

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Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.、 

Example 1:

Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.

Example 2:

Input: "aba"
Output: False

Example 3:

Input: "abcabcabcabc"
Output: True
Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)

---

給定一個非空的字符串，判斷它是否能夠由它的一個子串重複屢次構成。給定的字符串只含有小寫英文字母，而且長度不超過10000。

示例 1:

輸入: "abab"

輸出: True

解釋: 可由子字符串 "ab" 重複兩次構成。

示例 2:

輸入: "aba"

輸出: False

示例 3:

輸入: "abcabcabcabc"

輸出: True

解釋: 可由子字符串 "abc" 重複四次構成。 (或者子字符串 "abcabc" 重複兩次構成。)

---

148ms

 1 class Solution {
 2     //kmp算法
 3     func repeatedSubstringPattern(_ s: String) -> Bool {
 4         var arr:[Character] = [Character]()
 5         for char in s.characters
 6         {
 7             arr.append(char)
 8         }
 9         var i:Int = 1
10         var j:Int = 0
11         var n:Int = s.count
12         var dp:[Int] = [Int](repeating:0,count:n + 1)
13         while(i < n)
14         {
15             if arr[i] == arr[j]
16             {
17                 i += 1
18                 j += 1
19                 dp[i] = j
20             }
21             else if j == 0
22             {
23                 i += 1
24             }
25             else
26             {
27                 j = dp[j]
28             }
29         }
30         return dp[n] % (n - dp[n]) == 0 && dp[n] != 0
31     }
32 }

---

292ms

1 class Solution {
2     func repeatedSubstringPattern(_ s: String) -> Bool {
3         let ss = s + s
4         let str = ss[ss.index(after: ss.startIndex)..<ss.index(before: ss.endIndex)]
5         return str.contains(s)
6     }
7 }

---

480ms

 1 class Solution {
 2     func repeatedSubstringPattern(_ s: String) -> Bool {
 3         let length = s.count
 4         var index  = length / 2
 5 
 6         while index >= 1 {
 7             if length % index == 0 {
 8                 let c = length / index
 9                 var current = ""
10                 
11                 for _ in 0..<c {
12                     
13                     let offset = s.index(s.startIndex, offsetBy: index)
14                     current += String(s[..<offset])
15 
16                 }
17                 if current == s {
18                     return true
19                 }
20 
21             }
22             index -= 1
23         }
24  
25         return false
26     }
27 }

---

500ms

1 class Solution {
2     func repeatedSubstringPattern(_ s: String) -> Bool {
3         let chas = [Character](s)
4         let res = String(chas[1...]) + String(chas[..<(chas.count-1)])
5         
6         return res.contains(s)
7     }
8 }

---

604ms

 1 class Solution {
 2     func repeatedSubstringPattern(_ s: String) -> Bool {
 3         let count = s.count
 4         var huff = count / 2
 5         while huff >= 1 {
 6             if count % huff == 0 {
 7                 let toIndex = s.index(s.startIndex, offsetBy: huff)
 8                 let subString = s[s.startIndex..<toIndex]
 9                 
10                 var num = count / huff
11                 var sumString = ""
12                 
13                 while num > 0 {
14                     sumString = sumString + subString
15                     num = num - 1
16                 }
17                 
18                 if sumString == s {
19                     return true
20                 }
21             }
22             
23             huff = huff - 1
24         }
25         return false
26     }
27 }

---

3292ms

 1 class Solution {
 2     func repeatedSubstringPattern(_ s: String) -> Bool {
 3         let length = s.count
 4         
 5         var result = false;
 6         for index in 1...length {
 7             // 整除則對比
 8             if length % (index) == 0 {
 9                 // 從0到index
10                 let character = s.prefix(index)
11                 let increment = index;
12                 var start = increment;
13                 
14                 var isEqual = false;
15                 while (start < length) {
16                     let begin = s.index(s.startIndex, offsetBy: start)
17                     let stop = s.index(s.startIndex, offsetBy: start + increment)
18                     let temp = s[begin..<stop]
19                     
20                     if (character == temp) {
21                         start += increment;
22                         isEqual = true;
23                         continue;
24                     } else {
25                         isEqual = false;
26                         break;
27                     }
28                 }
29                 result = isEqual;
30                 if isEqual {
31                     break;
32                 }
33             }
34         }
35         return result
36     }
37 }