【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=0; i<words.size(); ++i){
            int row=0;
            
            for(int k=0; k<3; ++k){
                if(rows[k].count((char)tolower(words[i][0])) > 0) row = k;
            }
            
            validWords.push_back(words[i]);
            for(int j=1; j<words[i].size(); ++j){
                if(rows[row].count((char)tolower(words[i][j])) == 0){
                    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 == 1)
            res.push_back(str);
    }
    
    return res;
}
    
};