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:
- You may use one character in the keyboard more than once.
- 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上簡潔一點：
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;
}
};