500. Keyboard Row

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.

我的答案：

class Solution {public:    vector<string> findWords(vector<string>& words) {        set<char> st1 {'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'};        set<char> st2 {'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'};        set<char> st3 {'z', 'x', 'c', 'v', 'b', 'n', 'm'};    /*    st1.insert('q');        st1.insert('w');        st1.insert('e');        st1.insert('r');        st1.insert('t');        st1.insert('y');        st1.insert('u');        st1.insert('i');        st1.insert('o');        st1.insert('p');        st2.insert('a');        st2.insert('s');        st2.insert('d');        st2.insert('f');        st2.insert('g');        st2.insert('h');        st2.insert('j');        st2.insert('k');        st2.insert('l');        st3.insert('z');        st3.insert('x');        st3.insert('c');        st3.insert('v');        st3.insert('b');        st3.insert('n');        st3.insert('m');*/        vector<string> rs;        for(int i = 0; i < words.size(); ++i){            int status = 0;            bool flag = true;            for(int j = 0; j < words[i].length(); ++j){                if(st1.find(words[i][j]) != st1.end()){                    if(status == 0 || status == 1){                        status = 1;                    }else{                        flag = false;                        break;                    }                }                if(st2.find(words[i][j]) != st2.end()){                    if(status == 0 || status == 2){                        status = 2;                    }else{                        flag = false;                        break;                    }                }                if(st3.find(words[i][j]) != st3.end()){                    if(status == 0 || status == 3){                        status = 3;                    }else{                        flag = false;                        break;                    }                }            }            if(flag){               rs.push_back(words[i]);             }        }        return rs;    }};

leetcode上简洁一点：

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;    }};

运行时间上，初始化的时候给set赋值比挨个insert要快不少（待研究）