给定一个文本txt [0..n-1]和一个模式pat [0..m-1]，编写一个函数search(char pat []，char txt [])打印所有出现的pat []及其txt []中的排列(或字谜)。您可以假设n> m。
预期时间复杂度为O(n)
例子：
1) Input: txt[] = "BACDGABCDA" pat[] = "ABCD"
Output: Found at Index 0
Found at Index 5
Found at Index 6
2) Input: txt[] = "AAABABAA" pat[] = "AABA"
Output: Found at Index 0
Found at Index 1
Found at Index 4我们强烈建议您单击此处并进行实践，然后再继续解决方案。
这个问题与标准模式搜索问题略有不同，这里我们也需要搜索字谜。因此，我们无法直接应用标准模式搜索算法，例如KMP，Rabin Karp，Boyer Moore等。
一个简单的想法是修改Rabin Karp算法。例如，我们可以将哈希值作为所有字符的ASCII值的总和，以大质数为模。对于文本的每个字符，我们可以将当前字符添加到哈希值，然后减去上一个窗口的第一个字符。该解决方案看起来不错，但与标准Rabin Karp一样，该解决方案的最坏情况时间复杂度为O(mn)。当所有哈希值都匹配并且我们一一匹配所有字符时，就会发生最坏的情况。
我们可以假定字母表大小是固定的，因为我们有在ASCII最多256个可能的字符通常是真下达到O(n)的时间复杂度。这个想法是使用两个count数组：
1)第一个计数数组存储模式中字符的频率。
2)第二个计数数组在当前文本窗口中存储字符的频率。
需要注意的重要一点是，比较两个计数数组的时间复杂度为O(1)，因为它们中的元素数量是固定的(与模式和文本大小无关)。以下是此算法的步骤。
1)将模式频率的计数存储在第一个计数数组countP []中。还将字符的频率计数存储在数组countTW []中的文本的第一个窗口中。
2)现在运行一个从i = M到N-1的循环。循环执行。
…..a)如果两个计数数组相同，则发现一个事件。
…..b)countTW []中文本的当前字符的增量计数
…..c)countWT []中前一个窗口中第一个字符的递减计数
3)上面的循环未检查最后一个窗口，因此请显式检查它。
以下是上述算法的实现。
C++
// C++ program to search all anagrams of a pattern in a text
#include
#include
#define MAX 256
using namespace std;
// This function returns true if contents of arr1[] and arr2[]
// are same, otherwise false.
bool compare(char arr1[], char arr2[])
{
for (int i=0; i Java
// Java program to search all anagrams
// of a pattern in a text
public class GFG
{
static final int MAX = 256;
// This function returns true if contents
// of arr1[] and arr2[] are same, otherwise
// false.
static boolean compare(char arr1[], char arr2[])
{
for (int i = 0; i < MAX; i++)
if (arr1[i] != arr2[i])
return false;
return true;
}
// This function search for all permutations
// of pat[] in txt[]
static void search(String pat, String txt)
{
int M = pat.length();
int N = txt.length();
// countP[]: Store count of all
// characters of pattern
// countTW[]: Store count of current
// window of text
char[] countP = new char[MAX];
char[] countTW = new char[MAX];
for (int i = 0; i < M; i++)
{
(countP[pat.charAt(i)])++;
(countTW[txt.charAt(i)])++;
}
// Traverse through remaining characters
// of pattern
for (int i = M; i < N; i++)
{
// Compare counts of current window
// of text with counts of pattern[]
if (compare(countP, countTW))
System.out.println("Found at Index " +
(i - M));
// Add current character to current
// window
(countTW[txt.charAt(i)])++;
// Remove the first character of previous
// window
countTW[txt.charAt(i-M)]--;
}
// Check for the last window in text
if (compare(countP, countTW))
System.out.println("Found at Index " +
(N - M));
}
/* Driver program to test above function */
public static void main(String args[])
{
String txt = "BACDGABCDA";
String pat = "ABCD";
search(pat, txt);
}
}
// This code is contributed by Sumit GhoshPython3
# Python program to search all
# anagrams of a pattern in a text
MAX=256
# This function returns true
# if contents of arr1[] and arr2[]
# are same, otherwise false.
def compare(arr1, arr2):
for i in range(MAX):
if arr1[i] != arr2[i]:
return False
return True
# This function search for all
# permutations of pat[] in txt[]
def search(pat, txt):
M = len(pat)
N = len(txt)
# countP[]: Store count of
# all characters of pattern
# countTW[]: Store count of
# current window of text
countP = [0]*MAX
countTW = [0]*MAX
for i in range(M):
(countP[ord(pat[i]) ]) += 1
(countTW[ord(txt[i]) ]) += 1
# Traverse through remaining
# characters of pattern
for i in range(M,N):
# Compare counts of current
# window of text with
# counts of pattern[]
if compare(countP, countTW):
print("Found at Index", (i-M))
# Add current character to current window
(countTW[ ord(txt[i]) ]) += 1
# Remove the first character of previous window
(countTW[ ord(txt[i-M]) ]) -= 1
# Check for the last window in text
if compare(countP, countTW):
print("Found at Index", N-M)
# Driver program to test above function
txt = "BACDGABCDA"
pat = "ABCD"
search(pat, txt)
# This code is contributed
# by Upendra Singh BartwalC#
// C# program to search all anagrams
// of a pattern in a text
using System;
class GFG
{
public const int MAX = 256;
// This function returns true if
// contents of arr1[] and arr2[]
// are same, otherwise false.
public static bool compare(char[] arr1,
char[] arr2)
{
for (int i = 0; i < MAX; i++)
{
if (arr1[i] != arr2[i])
{
return false;
}
}
return true;
}
// This function search for all
// permutations of pat[] in txt[]
public static void search(string pat,
string txt)
{
int M = pat.Length;
int N = txt.Length;
// countP[]: Store count of all
// characters of pattern
// countTW[]: Store count of current
// window of text
char[] countP = new char[MAX];
char[] countTW = new char[MAX];
for (int i = 0; i < M; i++)
{
(countP[pat[i]])++;
(countTW[txt[i]])++;
}
// Traverse through remaining
// characters of pattern
for (int i = M; i < N; i++)
{
// Compare counts of current window
// of text with counts of pattern[]
if (compare(countP, countTW))
{
Console.WriteLine("Found at Index " +
(i - M));
}
// Add current character to
// current window
(countTW[txt[i]])++;
// Remove the first character of
// previous window
countTW[txt[i - M]]--;
}
// Check for the last window in text
if (compare(countP, countTW))
{
Console.WriteLine("Found at Index " +
(N - M));
}
}
// Driver Code
public static void Main(string[] args)
{
string txt = "BACDGABCDA";
string pat = "ABCD";
search(pat, txt);
}
}
// This code is contributed
// by Shrikant1输出：
Found at Index 0
Found at Index 5
Found at Index 6