[Swift]LeetCode1156.单字符重复子串的最大长度|SwapForMaximumRepeatedSubstring ...

OStack程序员社区-中国程序员成长平台 › 门户 › 编程› 移动开发›Swift 教程

原作者: [db:作者] 来自: [db:来源] 收藏邀请

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（www.zengqiang.org）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

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.

如果字符串中的所有字符都相同，那么这个字符串是单字符重复的字符串。

给你一个字符串 text，你只能交换其中两个字符一次或者什么都不做，然后得到一些单字符重复的子串。返回其中最长的子串的长度。

示例 1：

输入：text = "ababa"
输出：3

示例 2：

输入：text = "aaabaaa"
输出：6

示例 3：

输入：text = "aaabbaaa"
输出：4

示例 4：

输入：text = "aaaaa"
输出：5

示例 5：

输入：text = "abcdef"
输出：1

提示：

1 <= text.length <= 20000
text 仅由小写英文字母组成。

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 }