最长公共字谜子序列
给定两个长度分别为n1和n2的字符串str1和str2 。问题是找到最长子序列的长度，该子序列以字谜的形式出现在两个字符串中。
注意：字符串仅包含小写字母。
例子：
Input : str1 = "abdacp", str2 = "ckamb"
Output : 3
Subsequence of str1 = abc
Subsequence of str2 = cab
OR
Subsequence of str1 = bac
Subsequence of str2 = cab
These are longest common anagram subsequences.
Input : str1 = "abbcfke", str2 = "fbaafbly"
Output : 4方法：创建两个哈希表，说freq1和freq2 。将str1的每个字符的频率存储在freq1中。同样，将str2的每个字符的频率存储在freq2中。初始化len = 0。现在，对于每个小写字母，从两个哈希表中找到其最低频率并将其累积到len 。
C++
// C++ implementation to find the length of the
// longest common anagram subsequence
#include
using namespace std;
#define SIZE 26
// function to find the length of the
// longest common anagram subsequence
int longCommomAnagramSubseq(char str1[], char str2[],
int n1, int n2)
{
// hash tables for storing frequencies of
// each character
int freq1[SIZE], freq2[SIZE];
memset(freq1, 0, sizeof(freq1));
memset(freq2, 0, sizeof(freq2));
int len = 0;
// calculate frequency of each character
// of 'str1[]'
for (int i = 0; i < n1; i++)
freq1[str1[i] - 'a']++;
// calculate frequency of each character
// of 'str2[]'
for (int i = 0; i < n2; i++)
freq2[str2[i] - 'a']++;
// for each character add its minimum frequency
// out of the two strings in 'len'
for (int i = 0; i < SIZE; i++)
len += min(freq1[i], freq2[i]);
// required length
return len;
}
// Driver program to test above
int main()
{
char str1[] = "abdacp";
char str2[] = "ckamb";
int n1 = strlen(str1);
int n2 = strlen(str2);
cout << "Length = "
<< longCommomAnagramSubseq(str1, str2, n1, n2);
return 0;
} Java
// Java implementation to find
// the length() of the longest
// common anagram subsequence
import java.io.*;
class GFG
{
static int SIZE = 26;
// function to find the
// length() of the longest
// common anagram subsequence
static int longCommomAnagramSubseq(String str1,
String str2,
int n1, int n2)
{
// hash tables for
// storing frequencies
// of each character
int []freq1 = new int[SIZE];
int []freq2 = new int[SIZE];
for(int i = 0; i < SIZE; i++)
{
freq1[i] = 0;
freq2[i] = 0;
}
int len = 0;
// calculate frequency
// of each character of
// 'str1[]'
for (int i = 0; i < n1; i++)
freq1[(int)str1.charAt(i) - (int)'a']++;
// calculate frequency
// of each character
// of 'str2[]'
for (int i = 0; i < n2; i++)
freq2[(int)str2.charAt(i) - (int)'a']++;
// for each character add
// its minimum frequency
// out of the two Strings
// in 'len'
for (int i = 0; i < SIZE; i++)
len += Math.min(freq1[i],
freq2[i]);
// required length()
return len;
}
// Driver Code
public static void main(String args[])
{
String str1 = "abdacp";
String str2 = "ckamb";
int n1 = str1.length();
int n2 = str2.length();
System.out.print("Length = " +
longCommomAnagramSubseq(str1, str2,
n1, n2));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)Python 3
# Python 3 implementation to find
# the length of the longest common
# anagram subsequence
SIZE = 26
# function to find the length of the
# longest common anagram subsequence
def longCommomAnagramSubseq(str1, str2,
n1, n2):
# List for storing frequencies
# of each character
freq1 = [0] * SIZE
freq2 = [0] * SIZE
l = 0
# calculate frequency of each
# character of 'str1[]'
for i in range(n1):
freq1[ord(str1[i]) -
ord('a')] += 1
# calculate frequency of each
# character of 'str2[]'
for i in range(n2) :
freq2[ord(str2[i]) -
ord('a')] += 1
# for each character add its
# minimum frequency out of
# the two strings in 'len'
for i in range(SIZE):
l += min(freq1[i], freq2[i])
# required length
return l
# Driver Code
if __name__ == "__main__":
str1 = "abdacp"
str2 = "ckamb"
n1 = len(str1)
n2 = len(str2)
print("Length = ",
longCommomAnagramSubseq(str1, str2,
n1, n2))
# This code is contributed by ita_cC#
// C# implementation to find
// the length of the longest
// common anagram subsequence
using System;
class GFG
{
static int SIZE = 26;
// function to find the
// length of the longest
// common anagram subsequence
static int longCommomAnagramSubseq(string str1,
string str2,
int n1, int n2)
{
// hash tables for
// storing frequencies
// of each character
int []freq1 = new int[SIZE];
int []freq2 = new int[SIZE];
for(int i = 0; i < SIZE; i++)
{
freq1[i] = 0;
freq2[i] = 0;
}
int len = 0;
// calculate frequency
// of each character of
// 'str1[]'
for (int i = 0; i < n1; i++)
freq1[str1[i] - 'a']++;
// calculate frequency
// of each character
// of 'str2[]'
for (int i = 0; i < n2; i++)
freq2[str2[i] - 'a']++;
// for each character add
// its minimum frequency
// out of the two strings
// in 'len'
for (int i = 0; i < SIZE; i++)
len += Math.Min(freq1[i],
freq2[i]);
// required length
return len;
}
// Driver Code
static void Main()
{
string str1 = "abdacp";
string str2 = "ckamb";
int n1 = str1.Length;
int n2 = str2.Length;
Console.Write("Length = " +
longCommomAnagramSubseq(str1, str2,
n1, n2));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)PHP
Javascript
输出：
Length = 3时间复杂度： O(n+m)。
辅助空间： O(1)。