[Swift]LeetCode1156. 單字符重複子串的最大長度 | Swap For Maximum Repeated Substring

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：爲敢（WeiGanTechnologies）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-ozyeonwx-kn.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

A string is a repeated character string if all characters in that string are the same.  For example, "cccc" is a repeated character string.

Return the largest L such that: after either swapping two characters in text or doing nothing, there exists a repeated character substring of text with length L.

 

Example 1:

Input: text = "ababa"
Output: 3

Example 2:

Input: text = "aaabaaa"
Output: 6

Example 3:

Input: text = "aaabbaaa"
Output: 4

Example 4:

Input: text = "aaaaa"
Output: 5

Example 5:

Input: text = "abcdef"
Output: 1 

Constraints:

• 1 <= text.length <= 20000
• text consist of lowercase English characters only.

---

若是字符串中的全部字符都相同，那麼這個字符串是單字符重複的字符串。

給你一個字符串 text，你只能交換其中兩個字符一次或者什麼都不作，而後獲得一些單字符重複的子串。返回其中最長的子串的長度。

 

示例 1：

輸入：text = "ababa"
輸出：3

示例 2：

輸入：text = "aaabaaa"
輸出：6

示例 3：

輸入：text = "aaabbaaa"
輸出：4

示例 4：

輸入：text = "aaaaa"
輸出：5

示例 5：

輸入：text = "abcdef"
輸出：1 

提示：

• 1 <= text.length <= 20000
• text 僅由小寫英文字母組成。

---

56ms

 1 class Solution {
 2     func maxRepOpt1(_ text: String) -> Int {
 3         let chars = Array(text)
 4         var idxdic = [Character: [Int]]()
 5         for i in chars.indices {
 6             let c = chars[i]
 7             var arr = idxdic[c, default: [Int]()]
 8             arr.append(i)
 9             idxdic[c] = arr
10         }
11         var ans = 0
12         for (_, arr) in idxdic {
13             ans = max(ans, lenthOfArr(arr))
14         }
15         return ans
16     }
17 
18     func lenthOfArr(_ arr: [Int]) -> Int {
19         if arr.count == 0 { return 0 }
20         var tuples = [(Int, Int)]()
21         var curStar = arr[0]
22         var curEnd = arr[0]
23         for i in 1..<arr.count {
24             if arr[i]-curEnd == 1 {
25                 curEnd = arr[i]
26             } else {
27                 tuples.append((curStar, curEnd))
28                 curStar = arr[i]
29                 curEnd = arr[i]
30             }
31         }
32         tuples.append((curStar, curEnd))
33         var tuplefirst = tuples[0]
34         var ans = tuplefirst.1 - tuplefirst.0 + 1
35         for i in 1..<tuples.count {
36             let curTuple = tuples[i]
37             if curTuple.0 - tuplefirst.1 == 2 {
38                 let lastLen = tuplefirst.1 - tuplefirst.0 + 1
39                 let curLen = curTuple.1 - curTuple.0 + 1
40                 ans = max(ans, lastLen+curLen)
41                 if i != tuples.count - 1 || i > 1 {
42                     ans = max(ans, lastLen+curLen+1)   
43                 } 
44             } else {
45                 let curLen = curTuple.1 - curTuple.0 + 1
46                 let lastLen = tuplefirst.1 - tuplefirst.0 + 1
47                 ans = max(ans, lastLen+1) 
48                 ans = max(ans, curLen+1)
49             }
50             tuplefirst = curTuple
51         }
52         return ans
53     }
54 }

---

120ms

 1 class Solution {
 2     func maxRepOpt1(_ text: String) -> Int {
 3         let characters = Array(text)
 4         var result = 0
 5         var hash = [Character:[Int]]()
 6         
 7         for (index, character) in characters.enumerated() {
 8             hash[character] = (hash[character] ?? []) + [index]
 9         }
10         
11         for (key, value) in hash {
12             var consecutiveCount = 1, previousConsecutiveCount = 0, maximum = 0
13             var i = 1
14             
15             while i < value.count {
16                 if value[i] == value[i-1] + 1 { consecutiveCount += 1 }
17                 else {
18                     previousConsecutiveCount = value[i] == value[i-1] + 2 ? consecutiveCount : 0
19                     consecutiveCount = 1
20                 }
21                 maximum = max(maximum, consecutiveCount + previousConsecutiveCount)
22                 i += 1
23             }
24             
25             maximum = max(maximum, consecutiveCount + previousConsecutiveCount)
26             result = max(result, maximum + (value.count > maximum ? 1 : 0))
27         }
28         
29         return result
30     }
31 }

---

Runtime: 144 ms

Memory Usage: 21.7 MB

 1 class Solution {
 2     func maxRepOpt1(_ text: String) -> Int {
 3         let arrT:[Character] = Array(text)
 4         let n:Int = text.count
 5         var best:Int = 0
 6         let alphabet:[Character] = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
 7         for c in alphabet
 8         {
 9             var count:Int = 0
10             for i in 0..<n
11             {
12                 if arrT[i] == c
13                 {
14                     count += 1
15                 }
16             }
17             if count == n
18             {
19                 best = n
20                 continue
21             }
22             for i in 0..<n
23             {
24                 if arrT[i] != c
25                 {
26                     var left:Int = i
27                     var right:Int = i
28                     while (left > 0 && arrT[left - 1] == c)
29                     {
30                         left -= 1
31                     }
32                     while (right < n - 1 && arrT[right + 1] == c)
33                     {
34                         right += 1
35                     }
36                     var combined:Int = right - left
37                     if combined == count
38                     {
39                         best = max(best, count)
40                     }
41                     else
42                     {
43                         best = max(best, combined + 1)
44                     }
45                 }
46             }
47         }
48         return best        
49     }
50 }