[Swift]LeetCode647. 回文子串 | Palindromic Substrings

发布时间：2020-12-14 05:03:09 所属栏目：百科来源：网络整理

导读：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: "ab

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","aa","aaa".?

Note:

The input string length won‘t exceed 1000.

给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。

具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被计为是不同的子串。

示例 1:

输入: "abc"
输出: 3
解释: 三个回文子串: "a","c".

示例 2:

输入: "aaa"
输出: 6
说明: 6个回文子串: "a","aaa".

注意:

输入的字符串长度不会超过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)
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,input,&result)
 7             checkPalindrom(i,i + 1,&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 }

（编辑：李大同）

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