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  • 【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=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;
        }
    };
    View Code

    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

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  • 原文地址:https://www.cnblogs.com/happyamyhope/p/10919093.html
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