给定大小为N的字符串S ，它由数字[0 – 9]和字符‘。’组成。 ，任务是打印可以通过按给定顺序按移动键盘获得的字符串。
注意： “。”表示输入时出现中断。
Below is the image to represent the characters associated with each number in the keypad.
例子：
Input: S = “234”
Output: ADG
Explanation:
Pressing the keys 2, 3, and 4 once gives the resultant string as “ADG”.
Input: S = “22.22”
Output: BB
Explanation:
Pressing the key 2 twice gives B, and then again pressing the key twice gives B. Therefore, the resultant string is “BB”.
方法：可以通过以下方式解决给定的问题：将移动键盘映射存储在数组中，然后遍历字符串S并将其转换为等效的字符串。请按照以下步骤解决问题：
- 初始化一个空字符串，用ans表示，以存储所需的结果。
- 将与每个键关联的字符串存储在移动小键盘中的nums []数组中，以使nums [i]表示按下数字i时的字符集。
- 使用变量i遍历给定的字符串S并执行以下步骤：
- 如果S [i]等于‘。 ，然后将i加1，然后继续进行下一个迭代。
- 否则，将变量cnt初始化为0以存储相同字符的计数。
- 迭代直到S [i]等于S [i + 1]，并在每次迭代中检查以下条件：
- 如果cnt等于2且S [i]为2、3、4、5、6或8，则退出循环，因为键：2、3、4、5、6和8包含相同的数字字符，即3。
- 如果cnt等于3并且S [i]为7或9，则退出循环，因为键：7和9包含相同数量的字符，即4。
- 将cnt和i的值增加1。
- 如果S [i]是7或9，则将字符nums [str [i]] [cnt％4]添加到字符串ans 。
- 否则，将字符nums [str [i]] [cnt％3]添加到字符串ans 。
- 将i的值增加1。
- 完成上述步骤后，输出值字符串ans作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
void printSentence(string str)
{
// Store the mobile keypad mappings
char nums[][5]
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV",
"WXYZ" };
// Traverse the string str
int i = 0;
while (str[i] != '\0') {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (str[i + 1]
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2'
&& str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (str[i] == '\0')
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
cout << nums[str[i] - 48][count % 4];
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
cout << nums[str[i] - 48][count % 3];
}
i++;
}
}
// Driver Code
int main()
{
string str = "234";
printSentence(str);
return 0;
} Java
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
static void printSentence(String S)
{
// Store the mobile keypad mappings
String nums[]
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV", "WXYZ" };
char str[] = S.toCharArray();
// Traverse the string str
int i = 0;
while (i < str.length) {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (i + 1 < str.length
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2' && str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (i == str.length)
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
System.out.print(
nums[str[i] - 48].charAt(count % 4));
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
System.out.print(
nums[str[i] - 48].charAt(count % 3));
}
i++;
}
}
// Driver Code
public static void main(String[] args)
{
String str = "234";
printSentence(str);
}
}
// This code is contributed by Kingash.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
static void printSentence(string S)
{
// Store the mobile keypad mappings
string[] nums
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV", "WXYZ" };
char[] str = S.ToCharArray();
// Traverse the string str
int i = 0;
while (i < str.Length) {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (i + 1 < str.Length
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2' && str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (i == str.Length)
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
Console.Write(nums[str[i] - 48][count % 4]);
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
Console.Write(nums[str[i] - 48][count % 3]);
}
i++;
}
}
// Driver Code
public static void Main(string[] args)
{
string str = "234";
printSentence(str);
}
}
// This code is contributed by ukasp.输出：
ADG时间复杂度： O(N)
辅助空间： O(1)