【leetcode】500. Keyboard Row

问题描述：

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.

Example 1:

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

Note:

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

解法一：

常规解法，用unordered_set存储每一行的字母，依次寻找判断。

 class Solution {
 public:
     vector<string> findWords(vector<string>& words) {
        unordered_set<char> row1 {'q', 'w', 'e', 'r', 't', 'y','u', 'i', 'o', 'p'};
         unordered_set<char> row2 {'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'};
         unordered_set<char> row3 { 'z', 'x', 'c', 'v', 'b' ,'n', 'm'};
         vector<unordered_set<char>> rows {row1, row2, row3};

         vector<string> validWords;
         for(int i=; i<words.size(); ++i){
             int row=;

             for(int k=; k<; ++k){
                 if(rows[k].count((char)tolower(words[i][])) > ) row = k;
             }

             validWords.push_back(words[i]);
             for(int j=; j<words[i].size(); ++j){
                 if(rows[row].count((char)tolower(words[i][j])) == ){
                     validWords.pop_back();
                     break;
                 }
             }

         }
         return validWords;
     }
 };

解法二：

这种解法比较有启发性，看起来很有意思。

 class Solution {
 public:
 vector<string> findWords(vector<string>& words)
 {
     vector<string> res;

     for(auto str : words)
     {
         bool r1 = str.find_first_of("QWERTYUIOPqwertyuiop") == string::npos ? false : true;
         bool r2 = str.find_first_of("ASDFGHJKLasdfghjkl") == string::npos ? false : true;
         bool r3 = str.find_first_of("ZXCVBNMzxcvbnm") == string::npos ? false : true;

         if(r1 + r2 + r3 == )
             res.push_back(str);
     }

     return res;
 }

 };