Leetcode刷题21-500.键盘行(C++)

题目来源：链接: [https://leetcode-cn.com/problems/keyboard-row/].

1.问题描述

给定一个单词列表，只返回可以使用在键盘同一行的字母打印出来的单词。键盘如下图所示。

示例：

输入: ["Hello", "Alaska", "Dad", "Peace"]
输出: ["Alaska", "Dad"]

注意：

1. 你可以重复使用键盘上同一字符。
2. 你可以假设输入的字符串将只包含字母。

2.我的解决方案

一开始我的思路是对的，但是代码的执行力度不够。
主要还是基本功不扎实，要学会灵活运用 STL 标准库函数。
第二次的代码用了string 类的 find 函数。
第一次的代码好low啊，哈哈哈哈
第一次蠢蠢的代码如下：

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> res;
        int n1 = 0, n2 = 0, n3 = 0;
        for(int i = 0; i < word.size(); i++)
        {
            for(int j = 0; j < word[i].size(); j++)
            {
                if(word[i][j] == 'x' || word[i][j] == 'z' || word[i][j] == 'c' ||  word[i][j] == 'v' || word[i][j] == 'b' || word[i][j] == 'n' || word[i][j] == 'm' || word[i][j] == 'X' || word[i][j] == 'Z' || word[i][j] == 'C' ||  word[i][j] == 'V' || word[i][j] == 'B' || word[i][j] == 'N' || word[i][j] == 'M')
                {
                    n1++;
                }
                else if(word[i][j] == 'a' || word[i][j] == 's' || word[i][j] == 'd' ||  word[i][j] == 'f' || word[i][j] == 'g' || word[i][j] == 'h' || word[i][j] == 'j' || word[i][j] == 'k' || word[i][j] == 'l' || word[i][j] == 'A' ||  word[i][j] == 'S' || word[i][j] == 'D' || word[i][j] == 'F' || word[i][j] == 'G'|| word[i][j] == 'H'|| word[i][j] == 'J'|| word[i][j] == 'K'|| word[i][j] == 'L')
                {
                    n2++;
                }
                else
                {
                    n3++;
                }
            }
            
        }
    }
};

最终代码如下：

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> res;
        string s1 = "ZXCVBNMzxcvbnm";
        string s2 = "ASDFGHJKLasdfghjkl";
        string s3 = "QWERTYUIOPqwertyuiop";
        for(int i = 0; i < words.size(); i++)
        {
            int n1 = 0, n2 = 0, n3 = 0;
            for(int j = 0; j < words[i].size(); j++)
            {
               if(s1.find(words[i][j]) != -1)
               {
                   n1++;
               }
               if(s2.find(words[i][j]) != -1)
               {
                   n2++;
               }
               if(s3.find(words[i][j]) != -1)
               {
                   n3++;
               }
            }
            if(n1 == words[i].size() || n2 == words[i].size() || n3 == words[i].size())
            {
                res.push_back(words[i]);
            }
        }
        return res;
    }
};

3.大神们的解决方案

思路差不多，只是多了个大小写转换
大神1：

class Solution {

public:

    vector<string> findWords(vector<string>& words) {
        string L1 = "QWERTYUIOP";//"QWERTYUIOPqwertyuiop";
        string L2 = "ASDFGHJKL";//"ASDFGHJKLasdfghjkl";
        string L3 = "ZXCVBNM";//"ZXCVBNMzxcvbnm";
		vector<string> Fwords;
        for (int i = 0; i < words.size(); i++)
        {
            string strTemp(words[i]);
            int sum1 = 0, sum2 = 0, sum3 = 0;
            transform(words[i].begin(), words[i].end(), strTemp.begin(), ::toupper);
            for (int j = 0; j < strTemp.size(); j++)
            {
                if (L1.find(strTemp[j]) != -1)sum1++;
                if (L2.find(strTemp[j]) != -1)sum2++;
                if (L3.find(strTemp[j]) != -1)sum3++;
            }
            if (sum1 == strTemp.size()) Fwords.push_back(words[i]);
            if (sum2 == strTemp.size()) Fwords.push_back(words[i]);
            if (sum3 == strTemp.size()) Fwords.push_back(words[i]);
        }
	 	return Fwords;
        }
};

大神2：
感觉是用了哈希算法实现的， map 键值对。等我待研究。。。

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        string first1="QWERTYUIOP",second1="ASDFGHJKL",third1="ZXCVBNM";
        map <char,int> Hash;
        for(auto it : first1)
            Hash[it]=1,Hash[it+32]=1;
        for(auto it : second1)
            Hash[it]=2,Hash[it+32]=2;
        for(auto it : third1)
            Hash[it]=3,Hash[it+32]=3;
        vector <string> ans;
        for(auto it : words) {
            bool flag=1;
            if(it.size()==1) {
                ans.push_back(it);
                continue;
            }
            int temp=Hash[it[0]];
            for(int i=1;i<it.size();i++)
                if(Hash[it[i]]!=temp) {
                    flag=0;
                    break;
                }
            if(flag)
                ans.push_back(it);
        }
        return ans;
    }
};

4.我的收获

熟练掌握 STL 标准模板库啊啊 啊
加油！！~~~
2019/3/12 胡云层 于南京 21