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[Swift]LeetCode647. 回文子串 | Palindromic Substrings

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Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c". 

Example 2:

Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa". 

Note:

  1. The input string length won‘t exceed 1000.

給定一個字符串,你的任務是計算這個字符串中有多少個回文子串。

具有不同開始位置或結束位置的子串,即使是由相同的字符組成,也會被計為是不同的子串。

示例 1:

輸入: "abc"
輸出: 3
解釋: 三個回文子串: "a", "b", "c".

示例 2:

輸入: "aaa"
輸出: 6
說明: 6個回文子串: "a", "a", "a", "aa", "aa", "aaa".

註意:

  1. 輸入的字符串長度不會超過1000。

16ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3         var count = 0
 4         var s = Array(s)
 5         for i in 0..<s.count {
 6             if i + 1 < s.count, s[i] == s[i + 1
] { 7 count += numPalindromes(s, i, i + 1) 8 } 9 count += numPalindromes(s, i, i) 10 } 11 return count 12 } 13 14 func numPalindromes(_ s: [Character], _ i: Int, _ j: Int) -> Int { 15 var count = 0, i = i, j = j 16 while i >= 0, j < s.count, s[i] == s[j] { 17 i -= 1 18 j += 1 19 count += 1 20 } 21 return count 22 } 23 }

20ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3         var result = 0
 4         let input = Array(s)
 5         for i in 0..<input.count {
 6             checkPalindrom(i,i,input,&result)
 7             checkPalindrom(i,i + 1,input,&result)
 8         }
 9         return result
10     }
11     
12     func checkPalindrom(_ left: Int, _ right: Int, _ input: [Character], _ result: inout Int) {
13         let count = input.count
14         var i = left, j = right
15         while i >= 0, j < count {
16             if input[i] == input[j] {
17                 result += 1
18                 i -= 1
19                 j += 1
20             }else{
21                 break
22             }
23         }
24     }
25 }

28ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3         let arr = Array(s)        
 4         let cnt = s.count
 5         if cnt == 0 {
 6             return 0
 7         }        
 8         var result = 0
 9 
10         for center in 0..<cnt {
11             var left = center
12             var right = center
13             
14             while left >= 0 && right < cnt && arr[left] == arr[right] {
15                 result += 1
16                 left -= 1
17                 right += 1
18             }
19         }
20 
21         for center in 1..<cnt {
22             var left = center-1
23             var right = center
24             
25             while left >= 0 && right < cnt && arr[left] == arr[right] {
26                 result += 1
27                 left -= 1
28                 right += 1
29             }
30         }
31         return result
32     }    
33 }

32ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3         let s = Array(s)
 4         var n = s.count, ans = 0
 5         for center in 0..<2*n {
 6             var left = center / 2
 7             var right = left + center % 2
 8             while (left >= 0 && right < n && 
 9                    s[s.index(s.startIndex,offsetBy:left)] == 
10                    s[s.index(s.startIndex,offsetBy:right)]) {
11                 ans+=1
12                 left-=1
13                 right+=1
14             }
15         }
16         return ans
17     }
18 }

40ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3         let chars = Array(s)
 4         func extend(_ left: Int, _ right: Int) -> Int {
 5             var (left, right) = (left, right)
 6             var count = 0
 7             while left >= 0
 8                 && right < chars.count
 9                 && chars[left] == chars[right] {
10                     left -= 1
11                     right += 1
12                     count += 1
13                 }
14 
15             return count
16         }
17 
18         return chars.indices.map {
19             extend($0, $0) + extend($0, $0 + 1)
20         }
21         .reduce(0, +)
22     }
23 }

56ms

 1 class Solution {
 2     func countSubstrings(_ s: String) -> Int {
 3     guard s.count > 1 else{
 4         return 1
 5     }
 6     
 7     var manipulationStr : [Character] = []
 8     for char in Array(s){
 9         manipulationStr.append("#")
10         manipulationStr.append(char)
11     }
12     manipulationStr.append("#")
13     
14     var count : Int = 0
15     for movingIndex in 1..<manipulationStr.count - 1{
16         var leftIndex = movingIndex
17         var rightIndex = movingIndex
18         while leftIndex >= 0 && rightIndex < manipulationStr.count{
19             if manipulationStr[leftIndex] != manipulationStr[rightIndex]{
20                 break
21             }else{
22                 if manipulationStr[leftIndex] != "#"{
23                     count += 1
24                 }
25                 leftIndex -= 1
26                 rightIndex += 1
27             }
28         }
29     }
30     return count
31   }
32 }

[Swift]LeetCode647. 回文子串 | Palindromic Substrings