给定两个序列，打印它们中存在的所有最长子序列。
例子：
Input:
string X = "AGTGATG"
string Y = "GTTAG"
Output:
GTAG
GTTG
Input:
string X = "AATCC"
string Y = "ACACG"
Output:
ACC
AAC
Input:
string X = "ABCBDAB"
string Y = "BDCABA"
Output:
BCAB
BCBA
BDAB我们在这里讨论了最长公共子序列 (LCS) 问题。那里讨论的函数主要是找到 LCS 的长度。我们还讨论了如何在此处打印最长子序列。但是由于两个字符串的LCS 不是唯一的，在这篇文章中，我们将打印出 LCS 问题的所有可能解决方案。
以下是打印所有 LCS 的详细算法。
我们构建了 L[m+1][n+1] 表，如上一篇文章所述，并从 L[m][n] 开始遍历二维数组。对于矩阵中的当前单元格 L[i][j]，
a) 如果 X 和 Y 的最后一个字符相同(即 X[i-1] == Y[j-1])，则该字符必须出现在子串 X[0…i-1] 的所有 LCS 中，并且Y[0..j-1]。我们简单地对矩阵中的 L[i-1][j-1] 进行递归，并将当前字符附加到子串 X[0…i-2] 和 Y[0..j-2] 的所有可能的 LCS。
b) 如果 X 和 Y 的最后一个字符不相同(即 X[i-1] != Y[j-1])，则可以从矩阵的任一上侧构造 LCS(即 L[i-1 ][j]) 或从矩阵的左侧(即 L[i][j-1])取决于哪个值更大。如果两个值相等(即 L[i-1][j] == L[i][j-1])，那么它将从矩阵的两边构造。因此，基于 L[i-1][j] 和 L[i][j-1] 处的值，如果值相等，我们会朝更大的方向前进，或者朝两个方向前进。
以下是上述想法的递归实现 –
C++
/* Dynamic Programming implementation of LCS problem */
#include
using namespace std;
// Maximum string length
#define N 100
int L[N][N];
/* Returns set containing all LCS for X[0..m-1], Y[0..n-1] */
set findLCS(string X, string Y, int m, int n)
{
// construct a set to store possible LCS
set s;
// If we reaches end of either string, return
// a empty set
if (m == 0 || n == 0)
{
s.insert("");
return s;
}
// If the last characters of X and Y are same
if (X[m - 1] == Y[n - 1])
{
// recurse for X[0..m-2] and Y[0..n-2] in
// the matrix
set tmp = findLCS(X, Y, m - 1, n - 1);
// append current character to all possible LCS
// of substring X[0..m-2] and Y[0..n-2].
for (string str : tmp)
s.insert(str + X[m - 1]);
}
// If the last characters of X and Y are not same
else
{
// If LCS can be constructed from top side of
// the matrix, recurse for X[0..m-2] and Y[0..n-1]
if (L[m - 1][n] >= L[m][n - 1])
s = findLCS(X, Y, m - 1, n);
// If LCS can be constructed from left side of
// the matrix, recurse for X[0..m-1] and Y[0..n-2]
if (L[m][n - 1] >= L[m - 1][n])
{
set tmp = findLCS(X, Y, m, n - 1);
// merge two sets if L[m-1][n] == L[m][n-1]
// Note s will be empty if L[m-1][n] != L[m][n-1]
s.insert(tmp.begin(), tmp.end());
}
}
return s;
}
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int LCS(string X, string Y, int m, int n)
{
// Build L[m+1][n+1] in bottom up fashion
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = max(L[i - 1][j], L[i][j - 1]);
}
}
return L[m][n];
}
/* Driver program to test above function */
int main()
{
string X = "AGTGATG";
string Y = "GTTAG";
int m = X.length();
int n = Y.length();
cout << "LCS length is " << LCS(X, Y, m, n) << endl;
set s = findLCS(X, Y, m, n);
for (string str : s)
cout << str << endl;
return 0;
} Java
/* Dynamic Programming implementation of LCS problem */
import java.util.*;
class GFG
{
// Maximum String length
static int N = 100;
static int [][]L = new int[N][N];
/* Returns set containing all LCS for
X[0..m-1], Y[0..n-1] */
static Set findLCS(String X,
String Y, int m, int n)
{
// construct a set to store possible LCS
Set s = new HashSet<>();
// If we reaches end of either String,
// return a empty set
if (m == 0 || n == 0)
{
s.add("");
return s;
}
// If the last characters of X and Y are same
if (X.charAt(m - 1) == Y.charAt(n - 1))
{
// recurse for X[0..m-2] and Y[0..n-2]
// in the matrix
Set tmp = findLCS(X, Y, m - 1, n - 1);
// append current character to all possible LCS
// of subString X[0..m-2] and Y[0..n-2].
for (String str : tmp)
s.add(str + X.charAt(m - 1));
}
// If the last characters of X and Y are not same
else
{
// If LCS can be constructed from top side of
// the matrix, recurse for X[0..m-2] and Y[0..n-1]
if (L[m - 1][n] >= L[m][n - 1])
s = findLCS(X, Y, m - 1, n);
// If LCS can be constructed from left side of
// the matrix, recurse for X[0..m-1] and Y[0..n-2]
if (L[m][n - 1] >= L[m - 1][n])
{
Set tmp = findLCS(X, Y, m, n - 1);
// merge two sets if L[m-1][n] == L[m][n-1]
// Note s will be empty if L[m-1][n] != L[m][n-1]
s.addAll(tmp);
}
}
return s;
}
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
static int LCS(String X, String Y, int m, int n)
{
// Build L[m+1][n+1] in bottom up fashion
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i - 1) == Y.charAt(j - 1))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j],
L[i][j - 1]);
}
}
return L[m][n];
}
// Driver Code
public static void main(String[] args)
{
String X = "AGTGATG";
String Y = "GTTAG";
int m = X.length();
int n = Y.length();
System.out.println("LCS length is " +
LCS(X, Y, m, n));
Set s = findLCS(X, Y, m, n);
for (String str : s)
System.out.println(str);
}
}
// This code is contributed by 29AjayKumar Python3
# Dynamic Programming implementation of LCS problem
# Maximum string length
N = 100
L = [[0 for i in range(N)]
for j in range(N)]
# Returns set containing all LCS
# for X[0..m-1], Y[0..n-1]
def findLCS(x, y, m, n):
# construct a set to store possible LCS
s = set()
# If we reaches end of either string, return
# a empty set
if m == 0 or n == 0:
s.add("")
return s
# If the last characters of X and Y are same
if x[m - 1] == y[n - 1]:
# recurse for X[0..m-2] and Y[0..n-2] in
# the matrix
tmp = findLCS(x, y, m - 1, n - 1)
# append current character to all possible LCS
# of substring X[0..m-2] and Y[0..n-2].
for string in tmp:
s.add(string + x[m - 1])
# If the last characters of X and Y are not same
else:
# If LCS can be constructed from top side of
# the matrix, recurse for X[0..m-2] and Y[0..n-1]
if L[m - 1][n] >= L[m][n - 1]:
s = findLCS(x, y, m - 1, n)
# If LCS can be constructed from left side of
# the matrix, recurse for X[0..m-1] and Y[0..n-2]
if L[m][n - 1] >= L[m - 1][n]:
tmp = findLCS(x, y, m, n - 1)
# merge two sets if L[m-1][n] == L[m][n-1]
# Note s will be empty if L[m-1][n] != L[m][n-1]
for i in tmp:
s.add(i)
return s
# Returns length of LCS for X[0..m-1], Y[0..n-1]
def LCS(x, y, m, n):
# Build L[m+1][n+1] in bottom up fashion
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
L[i][j] = 0
elif x[i - 1] == y[j - 1]:
L[i][j] = L[i - 1][j - 1] + 1
else:
L[i][j] = max(L[i - 1][j],
L[i][j - 1])
return L[m][n]
# Driver Code
if __name__ == "__main__":
x = "AGTGATG"
y = "GTTAG"
m = len(x)
n = len(y)
print("LCS length is", LCS(x, y, m, n))
s = findLCS(x, y, m, n)
for i in s:
print(i)
# This code is contributed by
# sanjeev2552C#
// Dynamic Programming implementation
// of LCS problem
using System;
using System.Collections.Generic;
class GFG
{
// Maximum String length
static int N = 100;
static int [,]L = new int[N, N];
/* Returns set containing all LCS for
X[0..m-1], Y[0..n-1] */
static HashSet findLCS(String X,
String Y,
int m, int n)
{
// construct a set to store possible LCS
HashSet s = new HashSet();
// If we reaches end of either String,
// return a empty set
if (m == 0 || n == 0)
{
s.Add("");
return s;
}
// If the last characters of X and Y are same
if (X[m - 1] == Y[n - 1])
{
// recurse for X[0..m-2] and Y[0..n-2]
// in the matrix
HashSet tmp = findLCS(X, Y, m - 1, n - 1);
// append current character to all possible LCS
// of subString X[0..m-2] and Y[0..n-2].
foreach (String str in tmp)
s.Add(str + X[m - 1]);
}
// If the last characters of X and Y are not same
else
{
// If LCS can be constructed from top side of
// the matrix, recurse for X[0..m-2] and Y[0..n-1]
if (L[m - 1, n] >= L[m, n - 1])
s = findLCS(X, Y, m - 1, n);
// If LCS can be constructed from left side of
// the matrix, recurse for X[0..m-1] and Y[0..n-2]
if (L[m, n - 1] >= L[m - 1, n])
{
HashSet tmp = findLCS(X, Y, m, n - 1);
// merge two sets if L[m-1,n] == L[m,n-1]
// Note s will be empty if L[m-1,n] != L[m,n-1]
foreach (String str in tmp)
s.Add(str);
}
}
return s;
}
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
static int LCS(String X, String Y, int m, int n)
{
// Build L[m+1,n+1] in bottom up fashion
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i, j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i, j] = L[i - 1, j - 1] + 1;
else
L[i, j] = Math.Max(L[i - 1, j],
L[i, j - 1]);
}
}
return L[m, n];
}
// Driver Code
public static void Main(String[] args)
{
String X = "AGTGATG";
String Y = "GTTAG";
int m = X.Length;
int n = Y.Length;
Console.WriteLine("LCS length is " +
LCS(X, Y, m, n));
HashSet s = findLCS(X, Y, m, n);
foreach (String str in s)
Console.WriteLine(str);
}
}
// This code is contributed by Rajput-Ji Javascript
输出：
LCS length is 4
GTAG
GTTG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。