500. Keyboard Row

1. 思路
   将同一行的英文字符映射为Map中的同一值，只要比较每个String中的每个字符的Map映射值是否一致即可。

2. 代码

#include <iostream>
#include <map>
#include <vector>

using namespace std;

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> output;
        vector<char> line1 = {'Q','W','E','R','T','Y','U','I','O','P',
                              'q','w','e','r','t','y','u','i','o','p'
        };
        vector<char> line2 = {'A','S','D','F','G','H','J','K','L',
                              'a','s','d','f','g','h','j','k','l'
        };
        vector<char> line3 = {'Z','X','C','V','B','N','M',
                              'z','x','c','v','b','n','m'
        };
        
        map<char,int> charMap;
        for(int i = 0 ; i < line1.size() ; i++)
            charMap[line1[i]] = 1;
        for(int i = 0 ; i < line2.size() ; i++)
            charMap[line2[i]] = 2;
        for(int i = 0 ; i < line3.size() ; i++)
            charMap[line3[i]] = 3;
        
        for(int i = 0 ; i < words.size() ; i++) {
            int flag = charMap[words[i][0]];
            
            int j = 1;
            for( ; j < words[i].size() ; j++)
                if(charMap[words[i][j]] != flag)
                    break;
            if(j == words[i].size())
                output.push_back(words[i]);
        }
        return output;
    }
};

int main(int argc, const char * argv[]) {
    
    vector<vector<char>> board ;
    vector<string> input= {"Hello", "Alaska", "Dad", "Peace"};
    Solution s ;
    
    vector<string> output = s.findWords(input);
    for(int i = 0 ; i < output.size() ; i++)
        cout<<output[i]<<endl;
    
    return 0;
}