打印给定数组中的所有字符串，这些字符串可以使用 QWERTY 键盘的单行键来键入
原文:https://www . geeksforgeeks . org/print-从给定数组中打印所有字符串-可以使用单排 qwerty 键盘中的键进行键入/
给定一个由小写和大写字母组成的字符串组成的字符串数组 arr[] ，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 键盘的单行键键入。
示例:
输入:arr[]= {“Yeti”、“Had”、“GFG”、“comment”} 输出: Yeti Had GFG 解释: “Yeti”可从 1 st 行输入。 “Had”可以从第 2 行和第行输入。 “GFG”可以从 2 和行输入。 因此，需要的输出是雪人有 GFG 。
输入: arr[] = {“极客”，“为”，“极客”，“亲”} 输出:亲
方法:使用哈希可以解决问题。这个想法是遍历数组，对于每个字符串，检查该字符串的所有字符是否可以使用同一行的键键入。打印发现为真的字符串。按照以下步骤解决问题:
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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<string>& arr)
{
// Stores row number of all possible
// character of the strings
unordered_map<char, int> 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<string> words
= { "Yeti", "Had",
"GFG", "comment" };
findWordsSameRow(words);
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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<String> arr)
{
// Stores row number of all possible
// character of the strings
Map<Character,
Integer> mp = new HashMap<Character,
Integer>();
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<String> words = Arrays.asList(
"Yeti", "Had", "GFG", "comment" );
findWordsSameRow(words);
}
}
// This code is contributed by jithin
Python 3
# 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 avanitrachhadiya2155
C
// 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<string> arr)
{
// Stores row number of all possible
// character of the strings
Dictionary<char,
int> mp = new Dictionary<char,
int>();
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<string> words = new List<string>( new string[] {
"Yeti", "Had", "GFG", "comment" });
findWordsSameRow(words);
}
}
// This code is contributed by chitranayal
java 描述语言
<script>
// JavaScript 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
function findWordsSameRow(arr) {
// Stores row number of all possible
// character of the strings
var mp = {};
mp["q"] = 1;
mp["w"] = 1;
mp["e"] = 1;
mp["r"] = 1;
mp["t"] = 1;
mp["y"] = 1;
mp["u"] = 1;
mp["i"] = 1;
mp["o"] = 1;
mp["p"] = 1;
mp["a"] = 2;
mp["s"] = 2;
mp["d"] = 2;
mp["f"] = 2;
mp["g"] = 2;
mp["h"] = 2;
mp["j"] = 2;
mp["k"] = 2;
mp["l"] = 2;
mp["z"] = 3;
mp["x"] = 3;
mp["c"] = 3;
mp["v"] = 3;
mp["b"] = 3;
mp["n"] = 3;
mp["m"] = 3;
// Traverse the array
for (const word of 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
var flag = true;
// Stores row number of the first
// character of current string
var rowNum = mp[word[0].toLowerCase()];
// Stores length of word
var M = word.length;
// Traverse current string
for (var i = 1; i < M; i++) {
// If current character can't be
// typed using keys of rowNum only
if (mp[word[i].toLowerCase()] !== rowNum) {
// Update flag
flag = false;
break;
}
}
// If current string can be typed
// using keys from rowNum only
if (flag) {
// Print the string
document.write(word + " ");
}
}
}
}
// Driver Code
var words = ["Yeti", "Had", "GFG", "comment"];
findWordsSameRow(words);
</script>
Output:
Yeti Had GFG
时间复杂度: O(N * M)，其中 N 和 M 分别表示字符串的个数和最长字符串的长度。 辅助空间: O(26)
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