solve problem Keyboard Row

Author: P-ppc

README.md

@@ -9,4 +9,5 @@ All solutions will be accepted!
 |461|[Hamming Distance](https://leetcode-cn.com/problems/hamming-distance/description/)|[java/py/js](./algorithms/HammingDistance)|Easy|
 |476|[Number Complement](https://leetcode-cn.com/problems/number-complement/description/)|[java/py/js](./algorithms/NumberComplement)|Easy|
 |728|[Self Dividing Numbers](https://leetcode-cn.com/problems/self-dividing-numbers/description/)|[java/py/js](./algorithms/SelfDividingNumbers)|Easy|
-|804|[Unique Morse Code Words](https://leetcode-cn.com/problems/unique-morse-code-words/description/)|[java/py/js](./algorithms/UniqueMorseCodeWords)|Easy|
+|804|[Unique Morse Code Words](https://leetcode-cn.com/problems/unique-morse-code-words/description/)|[java/py/js](./algorithms/UniqueMorseCodeWords)|Easy|
+|500|[Keyboard Row](https://leetcode-cn.com/problems/keyboard-row/description/)|[java/py/js](./algorithms/KeyboardRow)|Easy|

algorithms/KeyboardRow/README.md

@@ -0,0 +1,2 @@
+# Keyboard Row
+we can use prime product to get modulo to solve this problem, but be careful about js's Number overflow!

algorithms/KeyboardRow/Solution.java

@@ -0,0 +1,36 @@
+class Solution {
+    public String[] findWords(String[] words) {
+        int[] primeList = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
+        long firstLineProduct = 1L;
+        long secondLineProduct = 1L;
+        long lastLineProduct = 1L;
+        for(char c : "qwertyuiop".toCharArray()) {
+            firstLineProduct *= primeList[(int) c - (int) 'a'];
+        }
+        for(char c : "asdfghjkl".toCharArray()) {
+            secondLineProduct *= primeList[(int) c - (int) 'a'];
+        }
+        for(char c : "zxcvbnm".toCharArray()) {
+            lastLineProduct *= primeList[(int) c - (int) 'a'];
+        }
+        List<String> result = new ArrayList<String>();
+        
+        for (String word : words) {
+            String unduplicatedWord = "";
+            for (String c : word.toLowerCase().split("")) {
+                if (!unduplicatedWord.contains(c)) {
+                    unduplicatedWord += c;
+                }
+            }
+            long product = 1L;
+            for(char c : unduplicatedWord.toCharArray()) {
+                product *= primeList[(int) c - (int) 'a'];
+            }
+            
+            if (firstLineProduct % product == 0 || secondLineProduct % product == 0 || lastLineProduct % product == 0) {
+                result.add(word);
+            }
+        }
+        return result.toArray(new String[result.size()]);  
+    }
+}

algorithms/KeyboardRow/solution.js

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+/**
+ * @param {string[]} words
+ * @return {string[]}
+ */
+var findWords = function(words) {
+    // js didn't support Long Integer, So we use another way to solve problem
+    let firstLine = 'qwertyuiop',
+        secondLine = 'asdfghjkl',
+        lastLine = 'zxcvbnm'
+    let array = []
+    words.forEach(word => {
+        let first = second = last = 0
+        Array.from(new Set(word.toLowerCase().split(''))).forEach(char => {
+            if (firstLine.indexOf(char) !== -1) first = 1
+            else if (secondLine.indexOf(char) !== -1) second = 1
+            else if (lastLine.indexOf(char) !== -1) last = 1
+        })
+        if (first + second + last === 1) {
+            array.push(word)
+        }
+    })
+    return array
+};

algorithms/KeyboardRow/solution.py

@@ -0,0 +1,19 @@
+class Solution(object):
+    def findWords(self, words):
+        """
+        :type words: List[str]
+        :rtype: List[str]
+        """
+        # convert to division
+        prime_list = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101]
+        first_line_product = reduce(lambda x, y : x * y, [prime_list[ord(x) - ord('a')] for x in list('qwertyuiop')])
+        second_line_product = reduce(lambda x, y : x * y, [prime_list[ord(x) - ord('a')] for x in list('asdfghjkl')])
+        last_line_product = reduce(lambda x, y : x * y, [prime_list[ord(x) - ord('a')] for x in list('zxcvbnm')])
+        
+        result_list = []
+        for word in words:
+            product = reduce(lambda x, y : x * y, [prime_list[ord(x) - ord('a')] for x in set(word.lower())])
+            if first_line_product % product == 0 or second_line_product % product == 0 or last_line_product % product == 0:
+                result_list.append(word)
+        return result_list
+