【leetcode】500_Keyboard Row

problem

500. Keyboard Row

题意：判断给出的某个单词是否可以使用键盘上的某一行字母type得到；

注意大小写的转换；

solution1: 使用set保存三行字符，查看每个字符所在的行数是否一致；

class Solution {

public:

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

        vector<string> ans;

        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'};

        for(int i=; i<words.size(); i++)

        {

            int one = , two =, three = ;

            for(auto ch : words[i])

            {

                if(ch<'a') ch += ;//

                if(row1.count(ch)) one = ;

                if(row2.count(ch)) two = ;

                if(row3.count(ch)) three = ;

                if(one+two+three>) break;

            }

            if(one+two+three==) ans.push_back(words[i]);

        }

        return ans;

    }

};

SET:

set containers are generally slower than unordered_set containers to access individual elements by their key, but they allow the direct iteration on subsets based on their order.

or:

判断每一行的字母数目是否与对应单词的长度一致；

class Solution {

public:

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

        vector<string> ans;

        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'};

        for(int i=; i<words.size(); i++)

        {

            int one = , two =, three = ;

            for(auto ch : words[i])

            {

                if(ch<'a') ch += ;//

                if(row1.count(ch)) one++;

                if(row2.count(ch)) two++;

                if(row3.count(ch)) three++;

            }

            int num = words[i].size();

            if(one==num || two==num || three==num) ans.push_back(words[i]);

        }

        return ans;

    }

};

solution2: 使用map映射键盘上每个字母所在的行数，判断某个单词每个字母所在行数是否一致；

参考

1. Leetcode_500. Keyboard Row;

2. GrandYang；

完