给定一个包含[0，9]中数字的整数数组，任务是打印数字可能代表的所有可能的字母组合。
遵循数字到字母的映射(就像在电话按钮上一样)。请注意， 0和1不会映射到任何字母。下图显示了所有映射：
例子：
Input: arr[] = {2, 3}
Output: ad ae af bd be bf cd ce cf
Input: arr[] = {9}
Output: w x y z
方法：现在让我们考虑一下如何以迭代方式解决该问题。递归解决方案是直观且通用的。我们继续递归地添加每个可能的字母，这将生成所有可能的字符串。
让我们考虑一下如何使用递归解决方案构建迭代解决方案。通过使用堆栈可以进行递归。因此，如果我们使用堆栈而不是递归函数，那将是一个迭代解决方案吗?从技术上讲，可以这样说，但是我们在逻辑上并没有做任何不同的事情。
堆栈是LIFO DS。我们可以使用其他数据结构吗?如果我们使用FIFO DS，会有什么区别?假设有一个队列。由于BFS是按队列完成的，而DFS是按堆栈完成的，两者之间有什么区别吗?
DFS和BFS之间的区别与此问题类似。在DFS中，我们将一一找到树中的每条路径。它将首先执行路径的所有步骤，而BFS一次将所有路径构建在一起。
因此，队列将完美地解决此问题。两种使用队列和堆栈的算法之间的唯一区别将是它们的形成方式。堆栈将完全一一形成所有字符串，而队列将一起形成所有字符串，即经过x次传递之后，所有字符串的长度将为x。
例如：
If the given number is "23",
then using queue, the letter combinations
obtained will be:
["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
and using stack, the letter combinations obtained will
be:
["cf","ce","cd","bf","be","bd","af","ae","ad"].
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return a vector that contains
// all the generated letter combinations
vector letterCombinationsUtil(const int number[],
int n,
const string table[])
{
// To store the generated letter combinations
vector list;
queue q;
q.push("");
while (!q.empty()) {
string s = q.front();
q.pop();
// If complete word is generated
// push it in the list
if (s.length() == n)
list.push_back(s);
else
// Try all possible letters for current digit
// in number[]
for (auto letter : table[number[s.length()]])
q.push(s + letter);
}
// Return the generated list
return list;
}
// Function that creates the mapping and
// calls letterCombinationsUtil
void letterCombinations(const int number[], int n)
{
// table[i] stores all characters that
// corresponds to ith digit in phone
string table[10]
= { "0", "1", "abc", "def", "ghi",
"jkl", "mno", "pqrs", "tuv", "wxyz" };
vector list
= letterCombinationsUtil(number, n, table);
// Print the contents of the vector
for (auto word : list)
cout << word << " ";
return;
}
// Driver code
int main()
{
int number[] = { 2, 3 };
int n = sizeof(number) / sizeof(number[0]);
// Function call
letterCombinations(number, n);
return 0;
} Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG {
// Function to return a vector that contains
// all the generated letter combinations
static ArrayList
letterCombinationsUtil(int[] number, int n,
String[] table)
{
// To store the generated letter combinations
ArrayList list = new ArrayList<>();
Queue q = new LinkedList<>();
q.add("");
while (!q.isEmpty()) {
String s = q.remove();
// If complete word is generated
// push it in the list
if (s.length() == n)
list.add(s);
else {
String val = table[number[s.length()]];
for (int i = 0; i < val.length(); i++)
{
q.add(s + val.charAt(i));
}
}
}
return list;
}
// Function that creates the mapping and
// calls letterCombinationsUtil
static void letterCombinations(int[] number, int n)
{
// table[i] stores all characters that
// corresponds to ith digit in phone
String[] table
= { "0", "1", "abc", "def", "ghi",
"jkl", "mno", "pqrs", "tuv", "wxyz" };
ArrayList list
= letterCombinationsUtil(number, n, table);
// Print the contents of the list
for (int i = 0; i < list.size(); i++) {
System.out.print(list.get(i) + " ");
}
}
// Driver code
public static void main(String args[])
{
int[] number = { 2, 3 };
int n = number.length;
// Funciton call
letterCombinations(number, n);
}
}
// This code is contributed by rachana soma Python
# Python3 implementation of the approach
from collections import deque
# Function to return a list that contains
# all the generated letter combinations
def letterCombinationsUtil(number, n, table):
list = []
q = deque()
q.append("")
while len(q) != 0:
s = q.pop()
# If complete word is generated
# push it in the list
if len(s) == n:
list.append(s)
else:
# Try all possible letters for current digit
# in number[]
for letter in table[number[len(s)]]:
q.append(s + letter)
# Return the generated list
return list
# Function that creates the mapping and
# calls letterCombinationsUtil
def letterCombinations(number, n):
# table[i] stores all characters that
# corresponds to ith digit in phone
table = ["0", "1", "abc", "def", "ghi", "jkl",
"mno", "pqrs", "tuv", "wxyz"]
list = letterCombinationsUtil(number, n, table)
s = ""
for word in list:
s += word + " "
print(s)
return
# Driver code
number = [2, 3]
n = len(number)
# Function call
letterCombinations(number, n)C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function to return a vector that contains
// all the generated letter combinations
static List
letterCombinationsUtil(int[] number, int n,
String[] table)
{
// To store the generated letter combinations
List list = new List();
Queue q = new Queue();
q.Enqueue("");
while (q.Count != 0) {
String s = q.Dequeue();
// If complete word is generated
// push it in the list
if (s.Length == n)
list.Add(s);
else {
String val = table[number[s.Length]];
for (int i = 0; i < val.Length; i++) {
q.Enqueue(s + val[i]);
}
}
}
return list;
}
// Function that creates the mapping and
// calls letterCombinationsUtil
static void letterCombinations(int[] number, int n)
{
// table[i] stores all characters that
// corresponds to ith digit in phone
String[] table
= { "0", "1", "abc", "def", "ghi",
"jkl", "mno", "pqrs", "tuv", "wxyz" };
List list
= letterCombinationsUtil(number, n, table);
// Print the contents of the list
for (int i = 0; i < list.Count; i++) {
Console.Write(list[i] + " ");
}
}
// Driver code
public static void Main(String[] args)
{
int[] number = { 2, 3 };
int n = number.Length;
// Function call
letterCombinations(number, n);
}
}
// This code is contributed by Princi Singh 输出
ad ae af bd be bf cd ce cf
想要从精选的最佳视频中学习并解决问题，请查看有关从基础到高级C++的C++基础课程以及有关语言和STL的C++ STL课程。要完成从学习语言到DS Algo等的更多准备工作，请参阅“完整面试准备课程” 。