【leetcode】500_Keyboard Row

problem

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=0; i<words.size(); i++)
        {
            int one = 0, two =0, three = 0;
            for(auto ch : words[i])
            {
                if(ch<'a') ch += 32;//
                if(row1.count(ch)) one = 1;
                if(row2.count(ch)) two = 1;
                if(row3.count(ch)) three = 1;
                if(one+two+three>1) break;
            }
            if(one+two+three==1) 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=0; i<words.size(); i++)
        {
            int one = 0, two =0, three = 0;
            for(auto ch : words[i])
            {
                if(ch<'a') ch += 32;//
                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;
    }
};