给定一个字符串数组arr[] ，由小写和大写字母组成的字符串组成，任务是打印给定数组中的所有字符串，这些字符串可以使用 QWERTY 键盘的单行键输入。
例子：
Input: arr[] = {“Yeti”, “Had”, “GFG”, “comment”}
Output: Yeti Had GFG
Explanation:
“Yeti” can be typed from the 1st row.
“Had” can be typed from the 2nd row.
“GFG” can be typed from the 2nd row.
Therefore, the required output is Yeti Had GFG.
Input: arr[] = {“Geeks”, “for”, “Geeks”, “Pro”}
Output: Pro
方法：可以使用Hashing解决问题。这个想法是遍历数组，对于每个字符串，检查是否可以使用同一行的键输入字符串的所有字符。打印发现它为真的字符串。请按照以下步骤解决问题：
- 初始化一个 Map，比如mp ，为每个字符存储键盘中出现该字符键的行号。
- 遍历数组并为每个字符串，检查字符串的所有字符在地图或没有分配给它相同的行数。如果发现为真，则打印字符串。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to print all strings that
// can be typed using keys of a single
// row in a QWERTY Keyboard
void findWordsSameRow(vector& arr)
{
// Stores row number of all possible
// character of the strings
unordered_map mp{
{ 'q', 1 }, { 'w', 1 }, { 'e', 1 }, { 'r', 1 },
{ 't', 1 }, { 'y', 1 }, { 'u', 1 }, { 'o', 1 },
{ 'p', 1 }, { 'i', 1 }, { 'a', 2 }, { 's', 2 },
{ 'd', 2 }, { 'f', 2 }, { 'g', 2 }, { 'h', 2 },
{ 'j', 2 }, { 'k', 2 }, { 'l', 2 }, { 'z', 3 },
{ 'x', 3 }, { 'c', 3 }, { 'v', 3 }, { 'b', 3 },
{ 'n', 3 }, { 'm', 3 }
};
// Traverse the array
for (auto word : arr) {
// If current string is
// not an empty string
if (!word.empty()) {
// Sets true / false if a string
// can be typed using keys of a
// single row or not
bool flag = true;
// Stores row number of the first
// character of current string
int rowNum
= mp[tolower(word[0])];
// Stores length of word
int M = word.length();
// Traverse current string
for (int i = 1; i < M; i++) {
// If current character can't be
// typed using keys of rowNum only
if (mp[tolower(word[i])]
!= rowNum) {
// Update flag
flag = false;
break;
}
}
// If current string can be typed
// using keys from rowNum only
if (flag) {
// Print the string
cout << word << " ";
}
}
}
}
// Driver Code
int main()
{
vector words
= { "Yeti", "Had",
"GFG", "comment" };
findWordsSameRow(words);
} Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to print all strings that
// can be typed using keys of a single
// row in a QWERTY Keyboard
static void findWordsSameRow(List arr)
{
// Stores row number of all possible
// character of the strings
Map mp = new HashMap();
mp.put('q', 1);
mp.put('w', 1);
mp.put('e', 1);
mp.put('r', 1);
mp.put('t', 1);
mp.put('y', 1);
mp.put('u', 1);
mp.put('i', 1);
mp.put('o', 1);
mp.put('p', 1);
mp.put('a', 2);
mp.put('s', 2);
mp.put('d', 2);
mp.put('f', 2);
mp.put('g', 2);
mp.put('h', 2);
mp.put('j', 2);
mp.put('k', 2);
mp.put('l', 2);
mp.put('z', 3);
mp.put('x', 3);
mp.put('c', 3);
mp.put('v', 3);
mp.put('b', 3);
mp.put('n', 3);
mp.put('m', 3);
// Traverse the array
for(String word : arr)
{
// If current string is
// not an empty string
if (word.length() != 0)
{
// Sets true / false if a string
// can be typed using keys of a
// single row or not
boolean flag = true;
// Stores row number of the first
// character of current string
int rowNum = mp.get(
Character.toLowerCase(word.charAt(0)));
// Stores length of word
int M = word.length();
// Traverse current string
for(int i = 1; i < M; i++)
{
// If current character can't be
// typed using keys of rowNum only
if (mp.get(Character.toLowerCase(
word.charAt(i))) != rowNum)
{
// Update flag
flag = false;
break;
}
}
// If current string can be typed
// using keys from rowNum only
if (flag)
{
// Print the string
System.out.print(word + " ");
}
}
}
}
// Driver Code
public static void main(String[] args)
{
List words = Arrays.asList(
"Yeti", "Had", "GFG", "comment" );
findWordsSameRow(words);
}
}
// This code is contributed by jithin Python3
# Python3 program to implement
# the above approach
# Function to print all strings that
# can be typed using keys of a single
# row in a QWERTY Keyboard
def findWordsSameRow(arr):
# Stores row number of all possible
# character of the strings
mp = { 'q' : 1, 'w' : 1, 'e' : 1, 'r' : 1,
't' : 1, 'y' : 1, 'u' : 1, 'o' : 1,
'p' : 1, 'i' : 1, 'a' : 2, 's' : 2,
'd' : 2, 'f' : 2, 'g' : 2, 'h' : 2,
'j' : 2, 'k' : 2, 'l' : 2, 'z' : 3,
'x' : 3, 'c' : 3, 'v' : 3, 'b' : 3,
'n' : 3, 'm' : 3 }
# Traverse the array
for word in arr:
# If current string is
# not an empty string
if (len(word) != 0):
# Sets true / false if a string
# can be typed using keys of a
# single row or not
flag = True
rowNum = mp[word[0].lower()]
# Stores length of word
M = len(word)
# Traverse current string
for i in range(1, M):
# If current character can't be
# typed using keys of rowNum only
if (mp[word[i].lower()] != rowNum):
# Update flag
flag = False
break
# If current string can be typed
# using keys from rowNum only
if (flag):
# Print the string
print(word, end = ' ')
# Driver Code
words = [ "Yeti", "Had", "GFG", "comment" ]
findWordsSameRow(words)
# This code is contributed by avanitrachhadiya2155C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to print all strings that
// can be typed using keys of a single
// row in a QWERTY Keyboard
static void findWordsSameRow(List arr)
{
// Stores row number of all possible
// character of the strings
Dictionary mp = new Dictionary();
mp.Add('q', 1);
mp.Add('w', 1);
mp.Add('e', 1);
mp.Add('r', 1);
mp.Add('t', 1);
mp.Add('y', 1);
mp.Add('u', 1);
mp.Add('i', 1);
mp.Add('o', 1);
mp.Add('p', 1);
mp.Add('a', 2);
mp.Add('s', 2);
mp.Add('d', 2);
mp.Add('f', 2);
mp.Add('g', 2);
mp.Add('h', 2);
mp.Add('j', 2);
mp.Add('k', 2);
mp.Add('l', 2);
mp.Add('z', 3);
mp.Add('x', 3);
mp.Add('c', 3);
mp.Add('v', 3);
mp.Add('b', 3);
mp.Add('n', 3);
mp.Add('m', 3);
// Traverse the array
foreach(string word in arr)
{
// If current string is
// not an empty string
if (word.Length != 0)
{
// Sets true / false if a string
// can be typed using keys of a
// single row or not
bool flag = true;
// Stores row number of the first
// character of current string
int rowNum = mp[ char.ToLower(word[0])];
// Stores length of word
int M = word.Length;
// Traverse current string
for(int i = 1; i < M; i++)
{
// If current character can't be
// typed using keys of rowNum only
if (mp[Char.ToLower(word[i])] != rowNum)
{
// Update flag
flag = false;
break;
}
}
// If current string can be typed
// using keys from rowNum only
if (flag)
{
// Print the string
Console.Write(word + " ");
}
}
}
}
// Driver Code
public static void Main(String[] args)
{
List words = new List( new string[] {
"Yeti", "Had", "GFG", "comment" });
findWordsSameRow(words);
}
}
// This code is contributed by chitranayal 输出：
Yeti Had GFG时间复杂度： O(N * M)，其中 N 和 M 分别表示字符串的数量和最长字符串的长度。
辅助空间： O(26)
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