Leetcode刷题（2）

Given a List of words, returnthe words that can be typed using letters of alphabet ononly 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.

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1. 首先先对vector<string> &words中的string进行遍历

2. 将string分解为单个字母

char *dst = new char[255];

    int i; 

    for(i=0;i <=src.length();i++) 

        dst[i]=src[i]; 

    dst[i] = '\0';

     return dst;

 

3. 循环，看这个string中的字母是否都在一行，是则输出，不是则不输出

！一定要注意大小写的转换

代码部分

classSolution {

public:

    vector<string>findWords(vector<string>& words) {

        char l1[10] ={'q','w','e','r','t','y','u','i','o','p'};

        char l2[9] ={'a','s','d','f','g','h','j','k','l'};

        char l3[7] = {'z','x','c','v','b','n','m'};

        vector<string> outputs;

        bool a=false,b=false,c=false;

        int m=0,n=0,o=0;

        for(int it=0;it<words.size();it++) {

        

            string str = words[it];

          

            for(int i=0; i<str.size(); i++){

                for(int k=0; k<10; k++) {

                    if((char)tolower(str[i]) ==l1[k])

                       ++m;

               }

            }

            if(m == str.size()) a = true;

           

         

           

            for(int i=0; i<str.size(); i++){

                for(int k=0; k<9; k++) {

                    if((char)tolower(str[i]) ==l2[k])

                       ++n;

                 

                }

            }

            if(n == str.size()) b = true;

          

           

            for(int i=0; i<str.size(); i++){

                for(int k=0; k<7; k++) {

                    if((char)tolower(str[i]) ==l3[k])

                       ++o;

                      

                }

            }

            if(o == str.size()) c = true;

          

            if(a || b || c)

              {outputs.push_back(str);}

             

            m=n=o=0;

            a=b=c=false;

        }

        return outputs;

    }

};

 

 

参考代码

classSolution {

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;

    }

};

 

总结：参考代码是先将函数定位到某一行，然后进行性判别，如果有某个字母不是在这一行，就移除这个元素。