feat: add swift implementation to lcof2 problem: No.020 (#3008)

Author: klever34

lcof2/剑指 Offer II 020. 回文子字符串的个数/README.md

@@ -145,6 +145,40 @@ func countSubstrings(s string) (ans int) {
 }
 ```
 
+#### Swift
+
+```swift
+class Solution {
+    private var s: String = ""
+
+    func countSubstrings(_ s: String) -> Int {
+        var ans = 0
+        self.s = s
+        let length = s.count
+        for i in 0..<length {
+            ans += countPalindromes(i, i)
+            ans += countPalindromes(i, i + 1)
+        }
+        return ans
+    }
+
+    private func countPalindromes(_ i: Int, _ j: Int) -> Int {
+        var cnt = 0
+        var i = i
+        var j = j
+        let chars = Array(s)
+        
+        while i >= 0 && j < chars.count && chars[i] == chars[j] {
+            cnt += 1
+            i -= 1
+            j += 1
+        }
+        
+        return cnt
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

lcof2/剑指 Offer II 020. 回文子字符串的个数/Solution.swift

@@ -0,0 +1,29 @@
+class Solution {
+    private var s: String = ""
+
+    func countSubstrings(_ s: String) -> Int {
+        var ans = 0
+        self.s = s
+        let length = s.count
+        for i in 0..<length {
+            ans += countPalindromes(i, i)
+            ans += countPalindromes(i, i + 1)
+        }
+        return ans
+    }
+
+    private func countPalindromes(_ i: Int, _ j: Int) -> Int {
+        var cnt = 0
+        var i = i
+        var j = j
+        let chars = Array(s)
+        
+        while i >= 0 && j < chars.count && chars[i] == chars[j] {
+            cnt += 1
+            i -= 1
+            j += 1
+        }
+        
+        return cnt
+    }
+}