【LeetCode】500. Keyboard Row

【链接】：keyboard-row
【描述】：
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.
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简单来说就是找字符串，该字符串的每个字母都要在美式键盘的同一行即可。
【思路】第一种方法三个set 集合存储键盘布局，利用set的find功能查找是否满足条件。第二种方法看了一下讨论区，强大的python，居然有子集功能！
【代码】：

/***********************
【LeetCode】500. Keyboard Row
Author：herongwei
Time:2017/2/8 13:10
language：C++
http://blog.csdn.net/u013050857
***********************/
#pragma comment(linker,"/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
const int maxm = 55;
const LL MOD = 999999997;
int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
inline LL read(){
    int  c=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
    return c*f;
}
/*Input: ["Hello", "Alaska", "Dad", "Peace"]
//qwertyuiop
//asdfghjkl
//zxcvbnm
Output: ["Alaska", "Dad"]*/
class Solution1 {
public:
    vector<string> findWords(vector<string>& words) {
       int words_size=words.size();
       vector<string> ret;
        unordered_set <char> row1 = { 'q','Q','w','W','e','E','r','R','t','T','y','Y','u','U','i','I','o','O','p','P' };
        unordered_set <char> row2 = { 'a','A','s','S','d','D','f','F','g','G','h','H','j','J','k','K','l','L'};
        unordered_set <char> row3 = { 'z','Z','x','X','c','C','v','V','b','B','n','N','m','M'};

       for(auto &word:words){
          bool f1=f2=f3=true;
          for(auto &ch:word){
              if(f1){
                auto it=row1.find(ch);
                if(it==row1.end()) f1=false;
              }
              if(f2){
                auto it=row2.find(ch);
                if(it==row2.end()) f2=false;
              } 
              if(f3){
                auto it=row3.find(ch);
                if(it==row3.end()) f3=false;
              }
          }
          if(f1 || f2 || f3) ret.push_back(word);
       }
       return ret;
    }
};

class Solution2(object):
  def findWords(self, words):
    row1, row2, row3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm');
    ret = [];
    for word in words:
      w = set(word.lower());
      if w.issubset(row1) or w.issubset(row2) or w.issubset(row3):
        ret.append(word);
    return ret;