feat: add swift implementation to lcof2 problem: No.086 (#3471)

Author: klever34

lcof2/剑指 Offer II 086. 分割回文子字符串/README.md

@@ -285,6 +285,49 @@ public class Solution {
 }
 ```
 
+#### Swift
+
+```swift
+class Solution {
+    private var n: Int = 0
+    private var s: String = ""
+    private var f: [[Bool]] = []
+    private var t: [String] = []
+    private var ans: [[String]] = []
+
+    func partition(_ s: String) -> [[String]] {
+        n = s.count
+        self.s = s
+        f = Array(repeating: Array(repeating: true, count: n), count: n)
+        
+        let chars = Array(s)
+        
+        for i in stride(from: n - 1, through: 0, by: -1) {
+            for j in i + 1 ..< n {
+                f[i][j] = chars[i] == chars[j] && f[i + 1][j - 1]
+            }
+        }
+        
+        dfs(0)
+        return ans
+    }
+
+    private func dfs(_ i: Int) {
+        if i == n {
+            ans.append(t)
+            return
+        }
+        for j in i ..< n {
+            if f[i][j] {
+                t.append(String(s[s.index(s.startIndex, offsetBy: i)...s.index(s.startIndex, offsetBy: j)]))
+                dfs(j + 1)
+                t.removeLast()
+            }
+        }
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

lcof2/剑指 Offer II 086. 分割回文子字符串/Solution.swift

@@ -0,0 +1,38 @@
+class Solution {
+    private var n: Int = 0
+    private var s: String = ""
+    private var f: [[Bool]] = []
+    private var t: [String] = []
+    private var ans: [[String]] = []
+
+    func partition(_ s: String) -> [[String]] {
+        n = s.count
+        self.s = s
+        f = Array(repeating: Array(repeating: true, count: n), count: n)
+        
+        let chars = Array(s)
+        
+        for i in stride(from: n - 1, through: 0, by: -1) {
+            for j in i + 1 ..< n {
+                f[i][j] = chars[i] == chars[j] && f[i + 1][j - 1]
+            }
+        }
+        
+        dfs(0)
+        return ans
+    }
+
+    private func dfs(_ i: Int) {
+        if i == n {
+            ans.append(t)
+            return
+        }
+        for j in i ..< n {
+            if f[i][j] {
+                t.append(String(s[s.index(s.startIndex, offsetBy: i)...s.index(s.startIndex, offsetBy: j)]))
+                dfs(j + 1)
+                t.removeLast()
+            }
+        }
+    }
+}