给出了一个已编码的字符串，任务是对其进行解码。字符串编码的模式如下。
[sub_str] ==> The substring 'sub_str'
appears count times. 例子：
Input : str[] = "1[b]"
Output : b
Input : str[] = "2[ab]"
Output : abab
Input : str[] = "2[a2[b]]"
Output : abbabb
Input : str[] = "3[b2[ca]]"
Output : bcacabcacabcaca
这个想法是使用两个堆栈，一个用于整数，另一个用于字符。
现在，遍历字符串，
- 每当我们遇到任何数字时，都将其压入整数堆栈，如果是字母(a至z)或方括号('[‘)，则将其压入字符堆栈。
- 每当遇到任何尖括号(’]’)时，请从字符堆栈中弹出字符，直到在字符堆栈中找不到开括号('[‘)。另外，从整数堆栈中弹出顶部元素，例如n。现在，使一个字符串重复弹出的字符n次。现在，将字符串的所有字符压入堆栈。
以下是此方法的实现：
C++
// C++ program to decode a string recursively
// encoded as count followed substring
#include
using namespace std;
// Returns decoded string for 'str'
string decode(string str)
{
stack integerstack;
stack stringstack;
string temp = "", result = "";
// Traversing the string
for (int i = 0; i < str.length(); i++)
{
int count = 0;
// If number, convert it into number
// and push it into integerstack.
if (str[i] >= '0' && str[i] <='9')
{
while (str[i] >= '0' && str[i] <= '9')
{
count = count * 10 + str[i] - '0';
i++;
}
i--;
integerstack.push(count);
}
// If closing bracket ']', pop elemment until
// '[' opening bracket is not found in the
// character stack.
else if (str[i] == ']')
{
temp = "";
count = 0;
if (! integerstack.empty())
{
count = integerstack.top();
integerstack.pop();
}
while (! stringstack.empty() && stringstack.top()!='[' )
{
temp = stringstack.top() + temp;
stringstack.pop();
}
if (! stringstack.empty() && stringstack.top() == '[')
stringstack.pop();
// Repeating the popped string 'temo' count
// number of times.
for (int j = 0; j < count; j++)
result = result + temp;
// Push it in the character stack.
for (int j = 0; j < result.length(); j++)
stringstack.push(result[j]);
result = "";
}
// If '[' opening bracket, push it into character stack.
else if (str[i] == '[')
{
if (str[i-1] >= '0' && str[i-1] <= '9')
stringstack.push(str[i]);
else
{
stringstack.push(str[i]);
integerstack.push(1);
}
}
else
stringstack.push(str[i]);
}
// Pop all the elmenet, make a string and return.
while (! stringstack.empty())
{
result = stringstack.top() + result;
stringstack.pop();
}
return result;
}
// Driven Program
int main()
{
string str = "3[b2[ca]]";
cout << decode(str) << endl;
return 0;
} Java
// Java program to decode a string recursively
// encoded as count followed substring
import java.util.Stack;
class Test
{
// Returns decoded string for 'str'
static String decode(String str)
{
Stack integerstack = new Stack<>();
Stack stringstack = new Stack<>();
String temp = "", result = "";
// Traversing the string
for (int i = 0; i < str.length(); i++)
{
int count = 0;
// If number, convert it into number
// and push it into integerstack.
if (Character.isDigit(str.charAt(i)))
{
while (Character.isDigit(str.charAt(i)))
{
count = count * 10 + str.charAt(i) - '0';
i++;
}
i--;
integerstack.push(count);
}
// If closing bracket ']', pop elemment until
// '[' opening bracket is not found in the
// character stack.
else if (str.charAt(i) == ']')
{
temp = "";
count = 0;
if (!integerstack.isEmpty())
{
count = integerstack.peek();
integerstack.pop();
}
while (!stringstack.isEmpty() && stringstack.peek()!='[' )
{
temp = stringstack.peek() + temp;
stringstack.pop();
}
if (!stringstack.empty() && stringstack.peek() == '[')
stringstack.pop();
// Repeating the popped string 'temo' count
// number of times.
for (int j = 0; j < count; j++)
result = result + temp;
// Push it in the character stack.
for (int j = 0; j < result.length(); j++)
stringstack.push(result.charAt(j));
result = "";
}
// If '[' opening bracket, push it into character stack.
else if (str.charAt(i) == '[')
{
if (Character.isDigit(str.charAt(i-1)))
stringstack.push(str.charAt(i));
else
{
stringstack.push(str.charAt(i));
integerstack.push(1);
}
}
else
stringstack.push(str.charAt(i));
}
// Pop all the elmenet, make a string and return.
while (!stringstack.isEmpty())
{
result = stringstack.peek() + result;
stringstack.pop();
}
return result;
}
// Driver method
public static void main(String args[])
{
String str = "3[b2[ca]]";
System.out.println(decode(str));
}
} Python3
# Python program to decode a string recursively
# encoded as count followed substring
# Returns decoded string for 'str'
def decode(Str):
integerstack = []
stringstack = []
temp = ""
result = ""
i = 0
# Traversing the string
while i < len(Str):
count = 0
# If number, convert it into number
# and push it into integerstack.
if (Str[i] >= '0' and Str[i] <='9'):
while (Str[i] >= '0' and Str[i] <= '9'):
count = count * 10 + ord(Str[i]) - ord('0')
i += 1
i -= 1
integerstack.append(count)
# If closing bracket ']', pop elemment until
# '[' opening bracket is not found in the
# character stack.
elif (Str[i] == ']'):
temp = ""
count = 0
if (len(integerstack) != 0):
count = integerstack[-1]
integerstack.pop()
while (len(stringstack) != 0 and stringstack[-1] !='[' ):
temp = stringstack[-1] + temp
stringstack.pop()
if (len(stringstack) != 0 and stringstack[-1] == '['):
stringstack.pop()
# Repeating the popped string 'temo' count
# number of times.
for j in range(count):
result = result + temp
# Push it in the character stack.
for j in range(len(result)):
stringstack.append(result[j])
result = ""
# If '[' opening bracket, push it into character stack.
elif (Str[i] == '['):
if (Str[i-1] >= '0' and Str[i-1] <= '9'):
stringstack.append(Str[i])
else:
stringstack.append(Str[i])
integerstack.append(1)
else:
stringstack.append(Str[i])
i += 1
# Pop all the elmenet, make a string and return.
while len(stringstack) != 0:
result = stringstack[-1] + result
stringstack.pop()
return result
# Driven code
if __name__ == '__main__':
Str = "3[b2[ca]]"
print(decode(Str))
# This code is contributed by PranchalK.C#
// C# program to decode a string recursively
// encoded as count followed substring
using System;
using System.Collections.Generic;
class GFG
{
// Returns decoded string for 'str'
public static string decode(string str)
{
Stack integerstack = new Stack();
Stack stringstack = new Stack();
string temp = "", result = "";
// Traversing the string
for (int i = 0; i < str.Length; i++)
{
int count = 0;
// If number, convert it into number
// and push it into integerstack.
if (char.IsDigit(str[i]))
{
while (char.IsDigit(str[i]))
{
count = count * 10 + str[i] - '0';
i++;
}
i--;
integerstack.Push(count);
}
// If closing bracket ']', pop elemment
// until '[' opening bracket is not found
// in the character stack.
else if (str[i] == ']')
{
temp = "";
count = 0;
if (integerstack.Count > 0)
{
count = integerstack.Peek();
integerstack.Pop();
}
while (stringstack.Count > 0 &&
stringstack.Peek() != '[')
{
temp = stringstack.Peek() + temp;
stringstack.Pop();
}
if (stringstack.Count > 0 &&
stringstack.Peek() == '[')
{
stringstack.Pop();
}
// Repeating the popped string 'temo'
// count number of times.
for (int j = 0; j < count; j++)
{
result = result + temp;
}
// Push it in the character stack.
for (int j = 0; j < result.Length; j++)
{
stringstack.Push(result[j]);
}
result = "";
}
// If '[' opening bracket, push it
// into character stack.
else if (str[i] == '[')
{
if (char.IsDigit(str[i - 1]))
{
stringstack.Push(str[i]);
}
else
{
stringstack.Push(str[i]);
integerstack.Push(1);
}
}
else
{
stringstack.Push(str[i]);
}
}
// Pop all the elmenet, make a
// string and return.
while (stringstack.Count > 0)
{
result = stringstack.Peek() + result;
stringstack.Pop();
}
return result;
}
// Driver Code
public static void Main(string[] args)
{
string str = "3[b2[ca]]";
Console.WriteLine(decode(str));
}
}
// This code is contributed by Shrikant13 输出：
bcacabcacabcaca
<！-以上“ 3 [b2 [ca]]”代码的图示>
<！- 
>