[Swift]LeetCode647. 回文子串 | Palindromic Substrings
阿新 • • 發佈:2019-03-06
app 多少 ice 結束 lin [] diff pan this
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:
- 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:
- 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