LeetCode2-Keyboard Row-[Easy]

题目Description###

Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

American keyboard

Example 1:
Input: ["Hello", "Alaska", "Dad", "Peace"] Output: ["Alaska", "Dad"]

Note:
You may use one character in the keyboard more than once.
You may assume the input string will only contain letters of alphabet.

思路1：####

白痴思维，大致就是看word是否只属于某一行。

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        list1 = ['qwertyuiopQWERTYUIOP']
        list2 = ['asdfghjklASDFGHJKL']
        list3 = ['ZXCVBNMzxcvbnm']
        rList = []
        i = 0
        
        while( i< len(words) ):
            if words[i] <= list1 or words[i] <= list2 or words[i] <= list3:
                rList.append(words[i])
            else:
                continue    
            i=i+1
                        
        return rList
        ```
结果：超时
![](http://upload-images.jianshu.io/upload_images/395849-e862c0b1d0dc65cc.png?imageMogr2/auto-orient/strip%7CimageView2/2/w/1240)
为什么会超时呢？

####思路2：####
利用set的属性，然而还是超时。

class Solution(object):
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
row1 = set('qwertyuiopQWERTYUIOP')
row2 = set('asdfghjklASDFGHJKL')
row3 = set('zxcvbnmZXCVBNM')
rList = []
i = 0

    while( i< len(words) ):
        x = set(words[i])
        if x.issubset(row1) or x.issubset(row2) or x.issubset(row3):
            rList.append(words[i])
        else:
            continue    
        i=i+1
                    
    return rList

结果：
![](http://upload-images.jianshu.io/upload_images/395849-19efc5eaa6a9f083.png?imageMogr2/auto-orient/strip%7CimageView2/2/w/1240)
为什么会超时呢？

####思路3：####
参考discuss中**hotea**的方法，使用filter和匿名函数。

class Solution(object):
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
row1 = set('qwertyuiopQWERTYUIOP')
row2 = set('asdfghjklASDFGHJKL')
row3 = set('zxcvbnmZXCVBNM')
return filter(lambda x: set(x).issubset(row1) or set(x).issubset(row2) or set(x).issubset(row3), words)