Li Kou daily question - day 26 -496 Next larger element I

2022.6.24 Did you brush the questions today ？

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subject ：

Here's an array of strings words , Returns only those that can be used in American keyboard Words printed with letters on the same line . The keyboard is shown in the figure below .

American keyboard in ：

The first line consists of the characters "qwertyuiop" form .
The second line consists of characters "asdfghjkl" form .
The third line consists of the characters "zxcvbnm" form .

analysis ：

That is to say, I will give you some strings , You need to decide if you are in the same line , If so, return , Otherwise, do not return .

Ideas ： Use violence to solve , First determine the first letter of the string to determine which line it is on , Then judge whether the following characters belong to line change .

analysis ：

1. Violent solution

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string>vec;
        int row1 = 0;
        int flag = 0;
        int successflag = 0;
        
        for (auto i = 0; i < words.size(); i++)
        {
            for (auto& j : words[i])
            {
                if (flag == 0)
                {
                    if (j == 'q' || j == 'w' || j == 'e' || j == 'r' || j == 't' || j == 'y' || j == 'u' || j == 'i' || j == 'o' || j == 'p' || j == 'Q' || j == 'W' || j == 'E' || j == 'R' || j == 'T' || j == 'Y' || j == 'U' || j == 'I' || j == 'O' || j == 'P')
                    {
                        row1 = 1;
                        flag = 1;
                        successflag = 1;
                    }
                    else if (j == 'a' || j == 's' || j == 'd' || j == 'f' || j == 'g' || j == 'h' || j == 'j' || j == 'k' || j == 'l' || j == 'A' || j == 'S' || j == 'D' || j == 'F' || j == 'G' || j == 'H' || j == 'J' || j == 'K' || j == 'L')
                    {
                        row1 = 2;
                        flag = 1;
                        successflag = 1;
                    }
                    else
                    {
                        row1 = 3;
                        flag = 1;
                        successflag = 1;
                    }
                }
                else
                {
                    
                    if (row1==1&&(j == 'q' || j == 'w' || j == 'e' || j == 'r' || j == 't' || j == 'y' || j == 'u' || j == 'i' || j == 'o' || j == 'p' || j == 'Q' || j == 'W' || j == 'E' || j == 'R' || j == 'T' || j == 'Y' || j == 'U' || j == 'I' || j == 'O' || j == 'P' ))
                    {
                        successflag = 1;
                    }
                    else if ( row1 == 2&&(j == 'a' || j == 's' || j == 'd' || j == 'f' || j == 'g' || j == 'h' || j == 'j' || j == 'k' || j == 'l' || j == 'A' || j == 'S' || j == 'D' || j == 'F' || j == 'G' || j == 'H' || j == 'J' || j == 'K' || j == 'L'))
                    {
                        successflag = 1;
                    }
                    else if ( row1 == 3&&(j == 'z' || j == 'x' || j == 'c' || j == 'v' || j == 'b' || j == 'n' || j == 'm' || j == 'Z' || j == 'X' || j == 'C' || j == 'V' || j == 'B' || j == 'N' || j == 'M'  ))
                    {
                        successflag = 1;
                    }
                    else
                    {
                        successflag = 0;
                        break;
                    }
                }
                
            }
            if (successflag == 1)
            {
                vec.emplace_back(words[i]);
                flag = 0;
            }
            else
            {
                flag = 0;
            }
        }

        return vec;
    }
};

2. Hashtable

First save the characters of each line into three hash tables , Then we judge each string , Whether there is a relation with a line , The key value pair of the hash table is used here 1, Let's see if the hash table exists

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
   
        string one = "qwertyuiopQWERTYUIOP";
        string two = "asdfghjklASDFGHJKL";
        string three = "zxcvbnmZXCVBNM";
        unordered_map<char, int>m1, m2, m3;
        for (auto q : one)
        {
            m1[q]++;
        }
        for (auto q : two)
        {
            m2[q]++;
        }
        for (auto q : three)
        {
            m3[q]++;
        }
        vector<string>vec;
        for (auto word : words)
        {
            bool b1 = true;
            bool b2 = true;
            bool b3 = true;
            for (auto &c : word)
            {
                b1 &= m1[c];
                b2 &= m2[c];
                b3 &= m3[c];
            }
            if (b1 || b2 || b3)
            {
                vec.emplace_back(word);
            }
        }
        return vec;
    }
};