给定两个长度分别为N和M 的字符串S1和S2 ，由小写字母组成，任务是通过从字符串S1 中删除所有出现的字符串S2来找到可以将S1减少到的最小长度。
例子：
Input: S1 =”fffoxoxoxfxo”, S2 = “fox”;
Output: 3
Explanation:
By removing “fox” starting from index 2, the string modifies to “ffoxoxfxo”.
By removing “fox” starting from index 1, the string modifies to “foxfxo”.
By removing “fox” starting from index 0, the string modifies to “fxo”.
Therefore, the minimum length of string S1 after removing all occurrences of S2 is 3.
Input: S1 =”abcd”, S2 = “pqr”
Output: 4
方法：解决这个问题的思路是使用Stack Data Structure。请按照以下步骤解决给定的问题：
- 初始化一个堆栈，将字符串S1的字符存储在其中。
- 遍历给定的字符串S1并执行以下操作：
- 将当前字符压入堆栈。
- 如果堆栈的大小至少为 M ，则检查堆栈的顶部M 个字符是否形成字符串S2 。如果发现是真的，则删除那些字符。
- 完成上述步骤后，剩余的堆栈大小就是所需的最小长度。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum length
// to which string str can be reduced to
// by removing all occurences of string K
int minLength(string str, int N,
string K, int M)
{
// Initialize stack of characters
stack stackOfChar;
for (int i = 0; i < N; i++) {
// Push character into the stack
stackOfChar.push(str[i]);
// If stack size >= K.size()
if (stackOfChar.size() >= M) {
// Create empty string to
// store characters of stack
string l = "";
// Traverse the string K in reverse
for (int j = M - 1; j >= 0; j--) {
// If any of the characters
// differ, it means that K
// is not present in the stack
if (K[j] != stackOfChar.top()) {
// Push the elements
// back into the stack
int f = 0;
while (f != l.size()) {
stackOfChar.push(l[f]);
f++;
}
break;
}
// Store the string
else {
l = stackOfChar.top()
+ l;
// Remove top element
stackOfChar.pop();
}
}
}
}
// Size of stack gives the
// minimized length of str
return stackOfChar.size();
}
// Driver Code
int main()
{
string S1 = "fffoxoxoxfxo";
string S2 = "fox";
int N = S1.length();
int M = S2.length();
// Function Call
cout << minLength(S1, N, S2, M);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the minimum length
// to which String str can be reduced to
// by removing all occurences of String K
static int minLength(String str, int N,
String K, int M)
{
// Initialize stack of characters
Stack stackOfChar = new Stack();
for (int i = 0; i < N; i++)
{
// Push character into the stack
stackOfChar.add(str.charAt(i));
// If stack size >= K.size()
if (stackOfChar.size() >= M)
{
// Create empty String to
// store characters of stack
String l = "";
// Traverse the String K in reverse
for (int j = M - 1; j >= 0; j--)
{
// If any of the characters
// differ, it means that K
// is not present in the stack
if (K.charAt(j) != stackOfChar.peek())
{
// Push the elements
// back into the stack
int f = 0;
while (f != l.length())
{
stackOfChar.add(l.charAt(f));
f++;
}
break;
}
// Store the String
else
{
l = stackOfChar.peek()
+ l;
// Remove top element
stackOfChar.pop();
}
}
}
}
// Size of stack gives the
// minimized length of str
return stackOfChar.size();
}
// Driver Code
public static void main(String[] args)
{
String S1 = "fffoxoxoxfxo";
String S2 = "fox";
int N = S1.length();
int M = S2.length();
// Function Call
System.out.print(minLength(S1, N, S2, M));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
# Function to find the minimum length
# to which string str can be reduced to
# by removing all occurences of string K
def minLength(Str, N, K, M) :
# Initialize stack of characters
stackOfChar = []
for i in range(N) :
# Push character into the stack
stackOfChar.append(Str[i])
# If stack size >= K.size()
if (len(stackOfChar) >= M) :
# Create empty string to
# store characters of stack
l = ""
# Traverse the string K in reverse
for j in range(M - 1, -1, -1) :
# If any of the characters
# differ, it means that K
# is not present in the stack
if (K[j] != stackOfChar[-1]) :
# Push the elements
# back into the stack
f = 0
while (f != len(l)) :
stackOfChar.append(l[f])
f += 1
break
# Store the string
else :
l = stackOfChar[-1] + l
# Remove top element
stackOfChar.pop()
# Size of stack gives the
# minimized length of str
return len(stackOfChar)
# Driver code
S1 = "fffoxoxoxfxo"
S2 = "fox"
N = len(S1)
M = len(S2)
# Function Call
print(minLength(S1, N, S2, M))
# This code is contributed by divyeshrabadiya07C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the minimum length
// to which String str can be reduced to
// by removing all occurences of String K
static int minLength(String str, int N,
String K, int M)
{
// Initialize stack of characters
Stack stackOfChar = new Stack();
for (int i = 0; i < N; i++)
{
// Push character into the stack
stackOfChar.Push(str[i]);
// If stack size >= K.Count
if (stackOfChar.Count >= M)
{
// Create empty String to
// store characters of stack
String l = "";
// Traverse the String K in reverse
for (int j = M - 1; j >= 0; j--)
{
// If any of the characters
// differ, it means that K
// is not present in the stack
if (K[j] != stackOfChar.Peek())
{
// Push the elements
// back into the stack
int f = 0;
while (f != l.Length)
{
stackOfChar.Push(l[f]);
f++;
}
break;
}
// Store the String
else
{
l = stackOfChar.Peek()
+ l;
// Remove top element
stackOfChar.Pop();
}
}
}
}
// Size of stack gives the
// minimized length of str
return stackOfChar.Count;
}
// Driver Code
public static void Main(String[] args)
{
String S1 = "fffoxoxoxfxo";
String S2 = "fox";
int N = S1.Length;
int M = S2.Length;
// Function Call
Console.Write(minLength(S1, N, S2, M));
}
}
// This code is contributed by 29AjayKumar Javascript
输出：
3时间复杂度： O(N*M)
辅助空间： O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live