给定一个整数N ，任务是按照字典顺序以单词的形式打印从1到N的所有数字(N <100000)。
例子 ：
Input: N = 11
Output: eight, eleven, five, four, nine, one, seven, six, three, two
Explanation:
The numbers from 1 to N is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
Their respective representations in words are {one, two, three, four, five, six, seven, eight, nine, ten, eleven}.
Their correct lexicographical order is {eight, eleven, eleven, five, four, nine, one, seven, six, three, two}.
Input: N = 5
Output: five, four, one, three, two
Explanation:
The numbers from 1 to N is 1, 2, 3, 4, 5.
Their respective representations in words are {one, two, three, four, five}.
Their correct lexicographical order is {five, four, one, three, two}.
方法：请按照以下步骤解决问题：
- 初始化大小为N + 1的数组arr [] ，该数组将在每个索引处存储从1到N的每个索引的字符串表示形式。
- 将所有从1到N的数字转换为单词，并将其存储在其相应的索引处。
- 以升序对数组arr []进行排序。
- 打印数组arr []中存在的元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to convert a number to words
string convert_to_words(int n)
{
// Stores the digits
char num[1000];
string str = to_string(n);
strcpy(num, str.c_str());
char* arr_ptr = &num[0];
int len = strlen(arr_ptr);
string ans = "";
// Base cases
if (len == 0)
{
ans += "Empty String";
return ans;
}
// Stores strings of unit place
string single_digits[]
= { "zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine" };
// Stores strings for corner cases
string two_digits[]
= { "", "ten", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen",
"seventeen", "eighteen", "nineteen" };
// Stores strings for ten's place digits
string tens_multiple[] = {
"", "", "twenty", "thirty", "forty",
"fifty", "sixty", "seventy", "eighty", "ninety"
};
// Stores strings for powers of 10
string tens_power[] = { "hundred", "thousand" };
// If given number contains a single digit
if (len == 1)
{
ans += single_digits[num[0] - '0'];
return ans;
}
// Iterate over all the digits
int x = 0;
while (x < len)
{
// Represent first 2 digits in words
if (len >= 3)
{
if (num[x] - '0' != 0)
{
ans += single_digits[num[x] - '0'];
ans += " ";
ans += tens_power[len - 3];
ans += " ";
}
--len;
}
// Represent last 2 digits in words
else
{
// Explicitly handle corner cases [10, 19]
if (num[x] - '0' == 1)
{
int sum = num[x] - '0' + num[x] - '0';
ans += two_digits[sum];
return ans;
}
// Explicitly handle corner case 20
else if (num[x] - '0' == 2
&& num[x + 1] - '0' == 0)
{
ans += "twenty";
return ans;
}
// For rest of the two digit
// numbers i.e., 21 to 99
else
{
int i = (num[x] - '0');
if (i > 0)
{
ans += tens_multiple[i];
ans += " ";
}
else
ans += "";
++x;
if (num[x] - '0' != 0)
ans += single_digits[num[x] - '0'];
}
}
++x;
}
return "";
}
// Function to print all the numbers
// up to n in lexicographical order
static void lexNumbers(int n)
{
vector s;
// Convert all numbers in words
for (int i = 1; i <= n; i++)
{
s.push_back(convert_to_words(i));
}
// Sort all strings
sort(s.begin(), s.end());
vector ans;
for (int i = 0; i < n; i++)
ans.push_back(s[i]);
// Print answer
for (int i = 0; i < n - 1; i++)
cout << ans[i] << ", ";
cout << ans[n - 1];
}
// Driver Code
int main()
{
int n = 5;
lexNumbers(n);
return 0;
}
// This code is contributed by Dharanendra L V Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to convert a number to words
static String convert_to_words(int n)
{
// Stores the digits
char num[] = String.valueOf(n)
.toCharArray();
int len = num.length;
String ans = "";
// Base cases
if (len == 0) {
ans += "Empty String";
return ans;
}
// Stores strings of unit place
String[] single_digits = new String[] {
"zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine"
};
// Stores strings for corner cases
String[] two_digits = new String[] {
"", "ten", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen",
"seventeen", "eighteen", "nineteen"
};
// Stores strings for ten's place digits
String[] tens_multiple = new String[] {
"", "", "twenty", "thirty", "forty",
"fifty", "sixty", "seventy", "eighty", "ninety"
};
// Stores strings for powers of 10
String[] tens_power
= new String[] { "hundred", "thousand" };
// If given number contains a single digit
if (len == 1) {
ans += single_digits[num[0] - '0'];
return ans;
}
// Iterate over all the digits
int x = 0;
while (x < num.length) {
// Represent first 2 digits in words
if (len >= 3) {
if (num[x] - '0' != 0) {
ans += single_digits[num[x] - '0'];
ans += " ";
ans += tens_power[len - 3];
ans += " ";
}
--len;
}
// Represent last 2 digits in words
else {
// Explicitly handle corner cases [10, 19]
if (num[x] - '0' == 1) {
int sum = num[x] - '0' + num[x] - '0';
ans += two_digits[sum];
return ans;
}
// Explicitly handle corner case 20
else if (num[x] - '0' == 2
&& num[x + 1] - '0' == 0) {
ans += "twenty";
return ans;
}
// For rest of the two digit
// numbers i.e., 21 to 99
else {
int i = (num[x] - '0');
if (i > 0) {
ans += tens_multiple[i];
ans += " ";
}
else
ans += "";
++x;
if (num[x] - '0' != 0)
ans += single_digits[num[x] - '0'];
}
}
++x;
}
return "";
}
// Function to print all the numbers
// up to n in lexicographical order
static void lexNumbers(int n)
{
Vector s = new Vector();
// Convert all numbers in words
for (int i = 1; i <= n; i++) {
s.add(convert_to_words(i));
}
// Sort all strings
Collections.sort(s);
Vector ans
= new Vector();
for (int i = 0; i < n; i++)
ans.add(s.get(i));
// Print answer
for (int i = 0; i < n - 1; i++)
System.out.print(
ans.get(i) + ", ");
System.out.print(ans.get(n - 1));
}
// Driver Code
public static void main(String[] args)
{
int n = 5;
lexNumbers(n);
}
} Python3
# Python3 program for the
# above approach
# Function to convert a number
# to words
def convert_to_words(n):
# Stores the digits
num = str(n)
length = len(num)
ans = ""
# Base cases
if (length == 0):
ans += "Empty String"
return ans
# Stores strings of unit place
single_digits = ["zero", "one", "two",
"three", "four", "five",
"six", "seven", "eight", "nine"]
# Stores strings for corner cases
two_digits = ["", "ten", "eleven",
"twelve", "thirteen",
"fourteen", "fifteen",
"sixteen", "seventeen",
"eighteen", "nineteen"]
# Stores strings for ten's place digits
tens_multiple = ["", "", "twenty",
"thirty", "forty",
"fifty", "sixty",
"seventy", "eighty",
"ninety"]
# Stores strings for powers of 10
tens_power = ["hundred", "thousand"]
# If given number contains a
# single digit
if (length == 1):
ans += single_digits[ord(num[0]) -
ord('0')]
return ans
# Iterate over all the digits
x = 0
while (x < len(num)):
# Represent first 2 digits
# in words
if (length >= 3) :
if (num[x] - '0' != 0):
ans += single_digits[ord(num[x]) -
ord('0')]
ans += " "
ans += tens_power[len - 3]
ans += " "
length -= 1
# Represent last 2 digits in words
else :
# Explicitly handle corner
# cases[10, 19]
if (ord(num[x]) -
ord('0') == 1):
sum = (ord(num[x]) - ord('0' ) +
ord(num[x]) - ord('0'))
ans += two_digits[sum]
return ans
# Explicitly handle corner
# case 20
elif (ord(num[x]) -
ord('0') == 2 and
ord(num[x + 1]) -
ord('0') == 0):
ans += "twenty"
return ans
# For rest of the two digit
# numbers i.e., 21 to 99
else:
i = (ord(num[x]) -
ord('0'))
if (i > 0) :
ans += tens_multiple[i]
ans += " "
else:
ans += ""
x += 1
if (ord(num[x]) -
ord('0') != 0):
ans += single_digits[ord(num[x]) -
ord('0')]
x += 1
return ""
# Function to print all the numbers
# up to n in lexicographical order
def lexNumbers(n):
s = []
# Convert all numbers in
# words
for i in range(1, n + 1):
s.append(convert_to_words(i))
# Sort all strings
s.sort()
ans = []
for i in range(n):
ans.append(s[i])
# Print answer
for i in range(n - 1):
print(ans[i], end = ", ")
print(ans[n - 1], end = "")
# Driver Code
if __name__ == "__main__":
n = 5
lexNumbers(n)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to convert a number to words
static String convert_to_words(int n)
{
// Stores the digits
char []num = String.Join("", n).ToCharArray();
int len = num.Length;
String ans = "";
// Base cases
if (len == 0)
{
ans += "Empty String";
return ans;
}
// Stores strings of unit place
String[] single_digits = new String[]{
"zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine" };
// Stores strings for corner cases
String[] two_digits = new String[]{
"", "ten", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen",
"seventeen", "eighteen", "nineteen" };
// Stores strings for ten's place digits
String[] tens_multiple = new String[]{
"", "", "twenty", "thirty", "forty",
"fifty", "sixty", "seventy", "eighty", "ninety" };
// Stores strings for powers of 10
String[] tens_power = new String[]{ "hundred",
"thousand" };
// If given number contains a single digit
if (len == 1)
{
ans += single_digits[num[0] - '0'];
return ans;
}
// Iterate over all the digits
int x = 0;
while (x < num.Length)
{
// Represent first 2 digits in words
if (len >= 3)
{
if (num[x] - '0' != 0)
{
ans += single_digits[num[x] - '0'];
ans += " ";
ans += tens_power[len - 3];
ans += " ";
}
--len;
}
// Represent last 2 digits in words
else
{
// Explicitly handle corner cases [10, 19]
if (num[x] - '0' == 1)
{
int sum = num[x] - '0' +
num[x] - '0';
ans += two_digits[sum];
return ans;
}
// Explicitly handle corner case 20
else if (num[x] - '0' == 2 &&
num[x + 1] - '0' == 0)
{
ans += "twenty";
return ans;
}
// For rest of the two digit
// numbers i.e., 21 to 99
else
{
int i = (num[x] - '0');
if (i > 0)
{
ans += tens_multiple[i];
ans += " ";
}
else
ans += "";
++x;
if (num[x] - '0' != 0)
ans += single_digits[num[x] - '0'];
}
}
++x;
}
return "";
}
// Function to print all the numbers
// up to n in lexicographical order
static void lexNumbers(int n)
{
List s = new List();
// Convert all numbers in words
for(int i = 1; i <= n; i++)
{
s.Add(convert_to_words(i));
}
// Sort all strings
s.Sort();
List ans = new List();
for(int i = 0; i < n; i++)
ans.Add(s[i]);
// Print answer
for(int i = 0; i < n - 1; i++)
Console.Write(ans[i] + ", ");
Console.Write(ans[n - 1]);
}
// Driver Code
public static void Main(String[] args)
{
int n = 5;
lexNumbers(n);
}
}
// This code is contributed by Rajput-Ji 输出：
five, four, one, three, two时间复杂度： O(NlogN)其中N是给定的整数。
辅助空间： O(N)