Merge pull request #335 from soapyigu/Math

Add a solution to IP to CIDR

Author: soapyigu

Math/IPToCIDR.swift

@@ -0,0 +1,69 @@
+**
+ * Question Link: https://leetcode.com/problems/ip-to-cidr/
+ * Primary idea: Bit manipulation. Get the most right 1 and update steps based on it.
+ *
+ * Time Complexity: O(mlogn), Space Complexity: O(m)
+ * m is the length of the ip string
+ */
+
+class IPToCIDR {
+    private let BASE = 256
+    
+    func ipToCIDR(_ ip: String, _ n: Int) -> [String] {
+        var currentIPInt = ipToInt(ip)
+        var res = [String](), n = n
+        
+        while n > 0 {
+            // get the most right one bit
+            var step = currentIPInt & -currentIPInt
+            
+            if step == 0 {
+                step = Int(pow(Double(2), Double(32)))
+            }
+            
+            while step > n {
+                step /= 2
+            }
+            
+            res.append(IntToIP(currentIPInt, step))
+            
+            currentIPInt += step
+            n -= step
+        }
+        
+        return res
+    }
+    
+    private func ipToInt(_ ip: String) -> Int {
+        var x = 0
+        let strings = ip.split(separator: ".")
+        
+        for str in strings {
+            x = x * BASE + Int(str)!
+        }
+        
+        return x
+    }
+    
+    private func IntToIP(_ x: Int, _ step: Int) -> String {
+        var res = Array(""), x = x
+
+        for i in 0..<4 {
+            res.insert(contentsOf: String(x % BASE), at: 0)
+            if i != 3 {
+                res.insert(".", at: 0)
+            }
+            x /= BASE
+        }
+
+        var len = 33, step = step
+        while step > 0 {
+            len -= 1
+            step /= 2
+        }
+
+        res += "/\(len)"
+
+        return String(res)
+    }
+}