给定长度为N的字符串S ，该字符串由小写字符组成，这些小写字符包含重新排列的数字[0-9]英文表示形式，任务是按升序打印这些数字。
例子：
Input: S = “fviefuro”
Output: 45
Explanation: The given string can be reshuffled to “fourfive”, Therefore, the digits represented by the strings are 4 and 5.
Input: S = “owoztneoer”
Output: 012
Explanation: The given string can be reshuffled to get “zeroonetwo”, Therefore, the digits represented by the strings are 0, 1 and 2.
天真的方法：最简单的方法是生成给定字符串的所有排列，对于每个排列，检查是否有可能找到由字符串表示的有效数字。如果发现为真，则按升序打印数字集。
时间复杂度： O(N * N！)
辅助空间： O(1)
高效的方法：这个想法是基于观察到某些字符仅以一个数字出现。
In ‘zero’, character ‘z’ is unique.
In ‘two’, character ‘w’ is unique.
In ‘four’, character ‘u’ is unique.
In ‘six’, character ‘x’ is unique.
In ‘eight’, character ‘g’ is unique.
In ‘three’, character ‘h’ is unique since word “eight” having character ‘h’ has already been considered.
In ‘one’, character ‘o’ is unique since words having character ‘o’ have already been considered.
In ‘five’, character f’ is unique since word “four” having character ‘f’ has already been considered.
In ‘seven’, character ‘v’ is unique.
In ‘nine’, character ‘i’ is unique since words having character ‘i’ have already been considered.
请按照以下步骤解决问题：
- 初始化一个空字符串，ANS存储所需的结果。
- 将字符串中每个字符的频率存储在M中。
- 创建一个唯一字符到其对应字符串的映射。
- 遍历地图，然后执行以下步骤：
- 将与数字相对应的唯一字符存储在变量x中。
- 获取x在M中的出现，并将其存储在变量freq中。
- 将相应的数字，频率次数附加到ans 。
- 遍历x对应的单词，并以M的频率减少其字符的频率。
- 打印字符串， ans作为结果。
下面是上述方法的实现：
Python3
# Python program to implement the above approach
from collections import Counter
# Function to construct the original set of digits
# from the string in ascending order
def construct_digits(s):
# Store the unique characters
# corresponding to word and number
k = ["z", "w", "u", "x", "g",
"h", "o", "f", "v", "i"]
l = ["zero", "two", "four", "six", "eight",
"three", "one", "five", "seven", "nine"]
c = [0, 2, 4, 6, 8, 3, 1, 5, 7, 9]
# Store the required result
ans = []
# Store the frequency of
# each character of S
d = Counter(s)
# Traverse the unique characters
for i in range(len(k)):
# Store the count of k[i] in S
x = d.get(k[i], 0)
# Traverse the corresponding word
for j in range(len(l[i])):
# Decrement the frequency
# of characters by x
d[l[i][j]]-= x
# Append the digit x times to ans
ans.append(str(c[i])*x)
# Sort the digits in ascending order
ans.sort()
return "".join(ans)
# Driver Code
# Given string, s
s = "fviefuro"
# Function Call
print(construct_digits(s))C#
// C# program to implement the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to construct the original set of digits
// from the string in ascending order
static string construct_digits(string s)
{
// Store the unique characters
// corresponding to word and number
char[] k = { 'z', 'w', 'u', 'x', 'g',
'h', 'o', 'f', 'v', 'i' };
string[] l = { "zero", "two", "four", "six", "eight",
"three", "one", "five", "seven", "nine" };
int[] c = { 0, 2, 4, 6, 8, 3, 1, 5, 7, 9 };
// Store the required result
List ans = new List();
// Store the frequency of
// each character of S
Dictionary d = new Dictionary();
for(int i = 0; i < s.Length; i++)
{
if (!d.ContainsKey(s[i]))
d[s[i]] = 0;
d[s[i]] += 1;
}
// Traverse the unique characters
for(int i = 0; i < k.Length; i++)
{
// Store the count of k[i] in S
int x = 0;
if (d.ContainsKey(k[i]))
x = d[k[i]];
// Traverse the corresponding word
for(int j = 0; j < l[i].Length; j++)
{
// Decrement the frequency
// of characters by x
if (d.ContainsKey(l[i][j]))
d[l[i][j]] -= x;
}
// Append the digit x times to ans
ans.Add(((c[i]) * x).ToString());
}
// Sort the digits in ascending order
ans.Sort();
string str = (String.Join("", ans.ToArray()));
return str.Replace("0", "");
}
// Driver Code
public static void Main(string[] args)
{
// Given string, s
string s = "fviefuro";
// Function Call
Console.WriteLine(construct_digits(s));
}
}
// This code is contributed by ukasp 45时间复杂度： O(N * log(N))
辅助空间： O(N)