[String] Optimize syntax for Keyboard Row

soapyigu · soapyigu · commit b9e52e2df306 · 2017-08-27T16:33:18.000-07:00
diff --git a/String/KeyboardRow.swift b/String/KeyboardRow.swift
@@ -1,6 +1,6 @@
 /**
  * Question Link: https://leetcode.com/problems/keyboard-row/
- * Primary idea: Convert each row to set to determine the word is subset or not.
+ * Primary idea: Use filter to determine the word is subset or not.
  * 
  * Note: You can also use intersect() or union() functions to solve this problem.
  *
@@ -10,26 +10,14 @@
 
 class KeyboardRow {
     func findWords(_ words: [String]) -> [String] {
-        var res = [String]()
-  
-        let rowOne = Set("qwertyuiop".characters), rowTwo = Set("asdfghjkl".characters), rowThree = Set("zxcvbnm".characters)
-  
-        for word in words {
-            if isInRow(word, rowOne) || isInRow(word, rowTwo) || isInRow(word, rowThree) {
-                res.append(word)
-            }
-        } 
-  
-        return res
+        let rowOne = "qwertyuiop", rowTwo = "asdfghjkl", rowThree = "zxcvbnm"
+
+        return words.filter { word in rowOne.contains(word) || rowTwo.contains(word) || rowThree.contains(word) }
     }
-    
-    fileprivate func isInRow(_ word: String, _ row: Set<Character>) -> Bool {
-        for char in word.lowercased().characters {
-            if !row.contains(char) {
-                return false
-            }
+
+    extension String {
+        func contains(_ word: String) -> Bool {
+            return word.filter { c in !self.contains(c) }.characters.count == 0
         }
-  
-        return true
     }
 }
