leetcode 500. Keyboard Row 键盘行字符串的查找 + 想法不错

曾经终败给现在 2022-06-03 00:29 180阅读 0赞

Given a List of words, return the words that can be typed using letters of alphabet on only one row’s of American keyboard like the image below.

![这里写图片描述][SouthEast]  
American keyboard

Example 1:  
Input: \[“Hello”, “Alaska”, “Dad”, “Peace”\]  
Output: \[“Alaska”, “Dad”\]  
Note:  
You may use one character in the keyboard more than once.  
You may assume the input string will only contain letters of alphabet.

起初读了好几次都没搞清楚题意，后来才明白，题意是这样的：给出n个字符串，从而判断每个字符串中的字符石头来自美式键盘上的同一行，若来自同一行，返回该string。过程将键盘上的每行字符存储到相应的vector或者数组中，然后循环Input中的每个string，并且循环string中的每个char，从而进行比较。

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        vector<string> findWords(vector<string>& words) 
        {
            vector<string> res;
            set <char> row1 = { 'q','w','e','r','t','y','u','i','o','p' };
            set <char> row2 = { 'a','s','d','f','g','h','j','k','l' };
            set <char> row3 = { 'z','x','c','v','b','n','m' };
            for (string word : words)
            {
                bool d1 = true, d2 = true, d3 = true;
                for (char c : word)
                {
                    if (d1 == true)
                    {
                        if (row1.find(tolower(c)) == row1.end())
                            d1 = false;
                    }
                    if (d2 == true)
                    {
                        if (row2.find(tolower(c)) == row2.end())
                            d2 = false;
                    }
                    if (d3 == true)
                    {
                        if (row3.find(tolower(c)) == row3.end())
                            d3 = false;
                    }
                }
                if (d1 || d2 || d3)
                    res.push_back(word);
            }
            return res;
        }
    };

或者使用字符串的查找

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    
    
    class Solution
    {
    public:
        vector<string> findWords(vector<string>& words)
        {
            vector<string> res;
            string row1 = "qwertyuiop";
            string row2 = "asdfghjkl";
            string row3 = "zxcvbnm";
            for (string word : words)
            {
                bool d1 = true, d2 = true, d3 = true;
                for (char c : word)
                {
                    if (d1 == true)
                    {
                        if (row1.find(tolower(c)) == row1.npos)
                            d1 = false;
                    }
                    if (d2 == true)
                    {
                        if (row2.find(tolower(c)) == row2.npos)
                            d2 = false;
                    }
                    if (d3 == true)
                    {
                        if (row3.find(tolower(c)) == row3.npos)
                            d3 = false;
                    }
                }
                if (d1 || d2 || d3)
                    res.push_back(word);
            }
            return res;
        }
    };

[SouthEast]: /images/20220603/cf466fe87b8248abbb6fda60c81c33e4.png