给定两个由N和M 个字符组成的字符串X和Y ，任务是找到字符串X的最长子序列，它是字符串Y的子字符串。
例子：
Input: X = “ABCD”, Y = “ACDBDCD”
Output: ACD
Explanation:
“ACD” is longest subsequence of X which is substring of Y.
Input: X = A, Y = A
Output: A
朴素方法：解决给定问题的最简单方法是找到给定字符串X 的所有子序列，并在所有生成的子序列中打印该子序列，该子序列是最大长度且是Y的子字符串。
时间复杂度： O(N*M*2 N )
辅助空间： O(N)
高效的方法：上述方法也可以通过使用动态规划进行优化。我们的想法是创建一个二维数组，DP [] []的尺寸(N + 1)*(M + 1)和状态DP的[i] [j]是X [0,1]的子序列的最大长度，其是Y[0, j] 的子串。请按照以下步骤解决问题：
- 创建一个大小为N+1行和M+1列的二维数组dp[][] 。
- 用0初始化矩阵的第一行和第一列。
- 按如下方式填充所有剩余的行：
- 如果X[i – 1]的值等于Y[j – 1]的值，则将dp[i][j]的值更新为(1 + dp[i – 1][j – 1] ) 。
- 否则，将dp[i][j]的值更新为dp[i – 1][j] 。
- 通过迭代矩阵中的最后一行，将所需序列的最大长度存储在变量len 中，并将最大单元格值的行和列索引分别存储在变量i和j 中。
- 创建一个变量，比如res来存储结果字符串并从最大单元格值回溯。
- 迭代直到len的值大于0 ，执行以下步骤：
- 如果X[i – 1]的值等于Y[j – 1] 的值，则将X[i – 1]附加到res并将len 、 i和j的值减 1。
- 否则，将i的值减 1。
- 完成上述步骤后，打印字符串res作为结果。
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 longest
// subsequence that matches with
// the substring of other string
string longestSubsequence(string X, string Y)
{
// Stores the lengths of strings
// X and Y
int n = X.size();
int m = Y.size();
// Create a matrix
vector> mat(n + 1, vector(m + 1));
// Initialize the matrix
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < m + 1; j++)
{
if (i == 0 || j == 0)
mat[i][j] = 0;
}
}
// Fill all the remaining rows
for(int i = 1; i < n + 1; i++)
{
for(int j = 1; j < m + 1; j++)
{
// If the characters are equal
if (X[i - 1] == Y[j - 1])
{
mat[i][j] = 1 + mat[i - 1][j - 1];
}
// If not equal, then
// just move to next
// in subsequence string
else
{
mat[i][j] = mat[i - 1][j];
}
}
}
// Find maximum length of the
// longest subsequence matching
// substring of other string
int len = 0, col = 0;
// Iterate through the last
// row of matrix
for(int i = 0; i < m + 1; i++)
{
if (mat[n][i] > len)
{
len = mat[n][i];
col = i;
}
}
// Store the required string
string res = "";
int i = n;
int j = col;
// Backtrack from the cell
while (len > 0)
{
// If equal, then add the
// character to res string
if (X[i - 1] == Y[j - 1])
{
res = X[i - 1] + res;
i--;
j--;
len--;
}
else
{
i--;
}
}
// Return the required string
return res;
}
// Driver code
int main()
{
string X = "ABCD";
string Y = "ACDBDCD";
cout << (longestSubsequence(X, Y));
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
class GFG {
// Function to find the longest
// subsequence that matches with
// the substring of other string
public static String longestSubsequence(
String X, String Y)
{
// Stores the lengths of strings
// X and Y
int n = X.length();
int m = Y.length();
// Create a matrix
int[][] mat = new int[n + 1][m + 1];
// Initialize the matrix
for (int i = 0; i < n + 1; i++) {
for (int j = 0; j < m + 1; j++) {
if (i == 0 || j == 0)
mat[i][j] = 0;
}
}
// Fill all the remaining rows
for (int i = 1;
i < n + 1; i++) {
for (int j = 1;
j < m + 1; j++) {
// If the characters are equal
if (X.charAt(i - 1)
== Y.charAt(j - 1)) {
mat[i][j] = 1
+ mat[i - 1][j - 1];
}
// If not equal, then
// just move to next
// in subsequence string
else {
mat[i][j] = mat[i - 1][j];
}
}
}
// Find maximum length of the
// longest subsequence matching
// substring of other string
int len = 0, col = 0;
// Iterate through the last
// row of matrix
for (int i = 0; i < m + 1; i++) {
if (mat[n][i] > len) {
len = mat[n][i];
col = i;
}
}
// Store the required string
String res = "";
int i = n;
int j = col;
// Backtrack from the cell
while (len > 0) {
// If equal, then add the
// character to res string
if (X.charAt(i - 1)
== Y.charAt(j - 1)) {
res = X.charAt(i - 1) + res;
i--;
j--;
len--;
}
else {
i--;
}
}
// Return the required string
return res;
}
// Driver Code
public static void main(String args[])
{
String X = "ABCD";
String Y = "ACDBDCD";
System.out.println(
longestSubsequence(X, Y));
}
}Python3
# Python3 program for the above approach
# Function to find the longest
# subsequence that matches with
# the substring of other string
def longestSubsequence(X, Y):
# Stores the lengths of strings
# X and Y
n = len(X)
m = len(Y)
# Create a matrix
mat = [[0 for i in range(m + 1)]
for j in range(n + 1)]
# Initialize the matrix
for i in range(0, n + 1):
for j in range(0, m + 1):
if (i == 0 or j == 0):
mat[i][j] = 0
# Fill all the remaining rows
for i in range(1, n + 1):
for j in range(1, m + 1):
# If the characters are equal
if (X[i - 1] == Y[j - 1]):
mat[i][j] = 1 + mat[i - 1][j - 1]
# If not equal, then
# just move to next
# in subsequence string
else:
mat[i][j] = mat[i - 1][j]
# Find maximum length of the
# longest subsequence matching
# substring of other string
len1 = 0
col = 0
# Iterate through the last
# row of matrix
for i in range(0, m + 1):
if (mat[n][i] > len1):
len1 = mat[n][i]
col = i
# Store the required string
res = ""
i = n
j = col
# Backtrack from the cell
while (len1 > 0):
# If equal, then add the
# character to res string
if (X[i - 1] == Y[j - 1]):
res = X[i - 1] + res
i -= 1
j -= 1
len1 -= 1
else:
i -= 1
# Return the required string
return res
# Driver code
X = "ABCD"
Y = "ACDBDCD"
print(longestSubsequence(X, Y))
# This code is contributed by amreshkumar3C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the longest
// subsequence that matches with
// the substring of other string
static string longestSubsequence(string X, string Y)
{
int i, j;
// Stores the lengths of strings
// X and Y
int n = X.Length;
int m = Y.Length;
// Create a matrix
int [,]mat = new int[n + 1, m + 1];
// Initialize the matrix
for(i = 0; i < n + 1; i++)
{
for(j = 0; j < m + 1; j++)
{
if (i == 0 || j == 0)
mat[i,j] = 0;
}
}
// Fill all the remaining rows
for(i = 1; i < n + 1; i++)
{
for(j = 1; j < m + 1; j++)
{
// If the characters are equal
if (X[i - 1] == Y[j - 1])
{
mat[i, j] = 1 + mat[i - 1, j - 1];
}
// If not equal, then
// just move to next
// in subsequence string
else
{
mat[i, j] = mat[i - 1, j];
}
}
}
// Find maximum length of the
// longest subsequence matching
// substring of other string
int len = 0, col = 0;
// Iterate through the last
// row of matrix
for(i = 0; i < m + 1; i++)
{
if (mat[n,i] > len)
{
len = mat[n,i];
col = i;
}
}
// Store the required string
string res = "";
i = n;
j = col;
// Backtrack from the cell
while (len > 0)
{
// If equal, then add the
// character to res string
if (X[i - 1] == Y[j - 1])
{
res = X[i - 1] + res;
i--;
j--;
len--;
}
else
{
i--;
}
}
// Return the required string
return res;
}
// Driver Code
public static void Main()
{
string X = "ABCD";
string Y = "ACDBDCD";
Console.Write(longestSubsequence(X, Y));
}
}
// This code is contributed by bgangwar59Javascript
输出：
ACD时间复杂度： O(N*M)
辅助空间： O(N*M)
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。