[Swift]LeetCode647. 回文子串 | Palindromic Substrings

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10482491.html 
➤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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

---

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", "b", "c".

示例 2:

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

注意:

1. 输入的字符串长度不会超过1000。

---

16ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
        var count = 0
        var s = Array(s)
        for i in 0..<s.count {
            if i + 1 < s.count, s[i] == s[i + 1] {
                count += numPalindromes(s, i, i + 1)
            }
            count += numPalindromes(s, i, i)
        }
        return count
    }

    func numPalindromes(_ s: [Character], _ i: Int, _ j: Int) -> Int {
        var count = 0, i = i, j = j
        while i >= 0, j < s.count, s[i] == s[j] {
            i -= 1
            j += 1
            count += 1
        }
        return count
    }
}

---

20ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
        var result = 0
        let input = Array(s)
        for i in 0..<input.count {
            checkPalindrom(i,i,input,&result)
            checkPalindrom(i,i + 1,input,&result)
        }
        return result
    }
    
    func checkPalindrom(_ left: Int, _ right: Int, _ input: [Character], _ result: inout Int) {
        let count = input.count
        var i = left, j = right
        while i >= 0, j < count {
            if input[i] == input[j] {
                result += 1
                i -= 1
                j += 1
            }else{
                break
            }
        }
    }
}

---

28ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
        let arr = Array(s)        
        let cnt = s.count
        if cnt == 0 {
            return 0
        }        
        var result = 0

        for center in 0..<cnt {
            var left = center
            var right = center
            
            while left >= 0 && right < cnt && arr[left] == arr[right] {
                result += 1
                left -= 1
                right += 1
            }
        }

        for center in 1..<cnt {
            var left = center-1
            var right = center
            
            while left >= 0 && right < cnt && arr[left] == arr[right] {
                result += 1
                left -= 1
                right += 1
            }
        }
        return result
    }    
}

---

32ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
        let s = Array(s)
        var n = s.count, ans = 0
        for center in 0..<2*n {
            var left = center / 2
            var right = left + center % 2
            while (left >= 0 && right < n && 
                   s[s.index(s.startIndex,offsetBy:left)] == 
                   s[s.index(s.startIndex,offsetBy:right)]) {
                ans+=1
                left-=1
                right+=1
            }
        }
        return ans
    }
}

---

40ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
        let chars = Array(s)
        func extend(_ left: Int, _ right: Int) -> Int {
            var (left, right) = (left, right)
            var count = 0
            while left >= 0
                && right < chars.count
                && chars[left] == chars[right] {
                    left -= 1
                    right += 1
                    count += 1
                }

            return count
        }

        return chars.indices.map {
            extend($0, $0) + extend($0, $0 + 1)
        }
        .reduce(0, +)
    }
}

---

56ms

class Solution {
    func countSubstrings(_ s: String) -> Int {
    guard s.count > 1 else{
        return 1
    }
    
    var manipulationStr : [Character] = []
    for char in Array(s){
        manipulationStr.append("#")
        manipulationStr.append(char)
    }
    manipulationStr.append("#")
    
    var count : Int = 0
    for movingIndex in 1..<manipulationStr.count - 1{
        var leftIndex = movingIndex
        var rightIndex = movingIndex
        while leftIndex >= 0 && rightIndex < manipulationStr.count{
            if manipulationStr[leftIndex] != manipulationStr[rightIndex]{
                break
            }else{
                if manipulationStr[leftIndex] != "#"{
                    count += 1
                }
                leftIndex -= 1
                rightIndex += 1
            }
        }
    }
    return count
  }
}