leetcode 500. Keyboard Row

Description

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

My solution

基本思路为将单词的字母"分类", 如果跟前面的分类不一致了, 说明跨行了.

class Solution {
public:
    vector<string> findWords(vector<string> &words) {
        vector<string> res;
        string top = "qwertyuiopQWERTYUIOP";
        string mid = "asdfghjklASDFGHJKL";
        string bot = "zxcvbnmZXCVBNM";
        unordered_map<char, int> mp;
        for (auto s:top) mp[s] = 1;
        for (auto s:mid) mp[s] = 2;
        for (auto s:bot) mp[s] = 3;
        for (auto word:words) {
            bool flag = true;
            for (auto w:word) {
                if (mp[w] != mp[word[0]]) {
                    flag = false;
                    break;
                }
            }
            if (flag) res.push_back(word);
        }
        return res;
    }
};

Discuss

发现一种比较简洁的方式(当然, 利用了find_first_of内置函数):

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;
}
    
};

Reference

• leetcode 500. Keyboard Row
• find_first_of

Example

// string::find_first_of
#include <iostream>       // std::cout
#include <string>         // std::string
#include <cstddef>        // std::size_t

int main ()
{
  std::string str ("Please, replace the vowels in this sentence by asterisks.");
  std::size_t found = str.find_first_of("aeiou");
  while (found!=std::string::npos)
  {
    str[found]='*';
    found=str.find_first_of("aeiou",found+1);
  }

  std::cout << str << '\n';

  return 0;
}

Pl**s*, r*pl*c* th* v*w*ls *n th*s s*nt*nc* by *st*r*sks.