给定两个字符串X和Y，打印同时包含X和Y作为子序列的最短字符串。如果存在多个最短超序列，请打印其中的任何一个。
例子：
Input: X = "AGGTAB", Y = "GXTXAYB"
Output: "AGXGTXAYB" OR "AGGXTXAYB"
OR Any string that represents shortest
supersequence of X and Y
Input: X = "HELLO", Y = "GEEK"
Output: "GEHEKLLO" OR "GHEEKLLO"
OR Any string that represents shortest
supersequence of X and Y
我们在这里讨论了如何打印两个给定字符串的最短可能超序列长度。在这篇文章中，我们将打印最短的超序列。
我们已经在下面的算法中进行了讨论，以找到先前后
Let X[0..m-1] and Y[0..n-1] be two strings and m and be respective
lengths.
if (m == 0) return n;
if (n == 0) return m;
// If last characters are same, then add 1 to result and
// recur for X[]
if (X[m-1] == Y[n-1])
return 1 + SCS(X, Y, m-1, n-1);
// Else find shortest of following two
// a) Remove last character from X and recur
// b) Remove last character from Y and recur
else return 1 + min( SCS(X, Y, m-1, n), SCS(X, Y, m, n-1) );
下表显示了如果我们使用动态编程以自底向上的方式对字符串X =“ AGGTAB”和Y =“ GXTXAYB”进行求解，则上述算法所遵循的步骤。
使用DP解决方案矩阵，我们可以按照以下步骤轻松打印两个字符串的最短超序列–
We start from the bottom-right most cell of the matrix and
push characters in output string based on below rules-
1. If the characters corresponding to current cell (i, j)
in X and Y are same, then the character is part of shortest
supersequence. We append it in output string and move
diagonally to next cell (i.e. (i - 1, j - 1)).
2. If the characters corresponding to current cell (i, j)
in X and Y are different, we have two choices -
If matrix[i - 1][j] > matrix[i][j - 1],
we add character corresponding to current
cell (i, j) in string Y in output string
and move to the left cell i.e. (i, j - 1)
else
we add character corresponding to current
cell (i, j) in string X in output string
and move to the top cell i.e. (i - 1, j)
3. If string Y reaches its end i.e. j = 0, we add remaining
characters of string X in the output string
else if string X reaches its end i.e. i = 0, we add
remaining characters of string Y in the output string.
以下是上述想法的实现–
C++
/* A dynamic programming based C++ program print
shortest supersequence of two strings */
#include
using namespace std;
// returns shortest supersequence of X and Y
string printShortestSuperSeq(string X, string Y)
{
int m = X.length();
int n = Y.length();
// dp[i][j] contains length of shortest supersequence
// for X[0..i-1] and Y[0..j-1]
int dp[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// Below steps follow recurrence relation
if(i == 0)
dp[i][j] = j;
else if(j == 0)
dp[i][j] = i;
else if(X[i - 1] == Y[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}
// Following code is used to print shortest supersequence
// dp[m][n] stores the length of the shortest supersequence
// of X and Y
int index = dp[m][n];
// string to store the shortest supersequence
string str;
// Start from the bottom right corner and one by one
// push characters in output string
int i = m, j = n;
while (i > 0 && j > 0)
{
// If current character in X and Y are same, then
// current character is part of shortest supersequence
if (X[i - 1] == Y[j - 1])
{
// Put current character in result
str.push_back(X[i - 1]);
// reduce values of i, j and index
i--, j--, index--;
}
// If current character in X and Y are different
else if (dp[i - 1][j] > dp[i][j - 1])
{
// Put current character of Y in result
str.push_back(Y[j - 1]);
// reduce values of j and index
j--, index--;
}
else
{
// Put current character of X in result
str.push_back(X[i - 1]);
// reduce values of i and index
i--, index--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str.push_back(X[i - 1]);
i--, index--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str.push_back(Y[j - 1]);
j--, index--;
}
// reverse the string and return it
reverse(str.begin(), str.end());
return str;
}
// Driver program to test above function
int main()
{
string X = "AGGTAB";
string Y = "GXTXAYB";
cstr << printShortestSuperSeq(X, Y);
return 0;
} Java
/* A dynamic programming based Java program print
shortest supersequence of two strings */
class GFG {
// returns shortest supersequence of X and Y
static String printShortestSuperSeq(String X, String Y)
{
int m = X.length();
int n = Y.length();
// dp[i][j] contains length of
// shortest supersequence
// for X[0..i-1] and Y[0..j-1]
int dp[][] = new int[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// Below steps follow recurrence relation
if (i == 0)
{
dp[i][j] = j;
}
else if (j == 0)
{
dp[i][j] = i;
}
else if (X.charAt(i - 1) == Y.charAt(j - 1))
{
dp[i][j] = 1 + dp[i - 1][j - 1];
}
else
{
dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
}
// Following code is used to print
// shortest supersequence dp[m][n] s
// tores the length of the shortest
// supersequence of X and Y
int index = dp[m][n];
// string to store the shortest supersequence
String str = "";
// Start from the bottom right corner and one by one
// push characters in output string
int i = m, j = n;
while (i > 0 && j > 0)
{
// If current character in X and Y are same, then
// current character is part of shortest supersequence
if (X.charAt(i - 1) == Y.charAt(j - 1))
{
// Put current character in result
str += (X.charAt(i - 1));
// reduce values of i, j and index
i--;
j--;
index--;
}
// If current character in X and Y are different
else if (dp[i - 1][j] > dp[i][j - 1])
{
// Put current character of Y in result
str += (Y.charAt(j - 1));
// reduce values of j and index
j--;
index--;
}
else
{
// Put current character of X in result
str += (X.charAt(i - 1));
// reduce values of i and index
i--;
index--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str += (X.charAt(i - 1));
i--;
index--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str += (Y.charAt(j - 1));
j--;
index--;
}
// reverse the string and return it
str = reverse(str);
return str;
}
static String reverse(String input)
{
char[] temparray = input.toCharArray();
int left, right = 0;
right = temparray.length - 1;
for (left = 0; left < right; left++, right--)
{
// Swap values of left and right
char temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return String.valueOf(temparray);
}
// Driver code
public static void main(String[] args)
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println(printShortestSuperSeq(X, Y));
}
}
// This code is contributed by 29AjayKumarPython3
# A dynamic programming based Python3 program print
# shortest supersequence of two strings
# returns shortest supersequence of X and Y
def printShortestSuperSeq(x, y):
m = len(x)
n = len(y)
# dp[i][j] contains length of shortest
# supersequence for X[0..i-1] and Y[0..j-1]
dp = [[0 for i in range(n + 1)]
for j in range(n + 1)]
# Fill table in bottom up manner
for i in range(m + 1):
for j in range(n + 1):
# Below steps follow recurrence relation
if i == 0:
dp[i][j] = j
elif j == 0:
dp[i][j] = i
elif x[i - 1] == y[j - 1]:
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j],
dp[i][j - 1])
# Following code is used to print
# shortest supersequence
# dp[m][n] stores the length of the
# shortest supersequence of X and Y
index = dp[m][n]
# string to store the shortest supersequence
string = ""
# Start from the bottom right corner and
# one by one push characters in output string
i = m
j = n
while i > 0 and j > 0:
# If current character in X and Y are same,
# then current character is part of
# shortest supersequence
if x[i - 1] == y[j - 1]:
# Put current character in result
string += x[i - 1]
# reduce values of i, j and index
i -= 1
j -= 1
index -= 1
# If current character in X and Y are different
elif dp[i - 1][j] > dp[i][j - 1]:
# Put current character of Y in result
string += y[j - 1]
# reduce values of j and index
j -= 1
index -= 1
else:
# Put current character of X in result
string += x[i - 1]
# reduce values of i and index
i -= 1
index -= 1
# If Y reaches its end, put remaining characters
# of X in the result string
while i > 0:
string += x[i - 1]
i -= 1
index -= 1
# If X reaches its end, put remaining characters
# of Y in the result string
while j > 0:
string += y[j - 1]
j -= 1
index -= 1
string = list(string)
# reverse the string and return it
string.reverse()
return ''.join(string)
# Driver Code
if __name__ == "__main__":
x = "AGGTAB"
y = "GXTXAYB"
print(printShortestSuperSeq(x, y))
# This code is contributed by
# sanjeev2552C#
/* A dynamic programming based C# program print
shortest supersequence of two strings */
using System;
class GFG
{
// returns shortest supersequence of X and Y
static String printShortestSuperSeq(String X, String Y)
{
int m = X.Length;
int n = Y.Length;
// dp[i,j] contains length of
// shortest supersequence
// for X[0..i-1] and Y[0..j-1]
int [,]dp = new int[m + 1, n + 1];
int i, j;
// Fill table in bottom up manner
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
// Below steps follow recurrence relation
if (i == 0)
{
dp[i, j] = j;
}
else if (j == 0)
{
dp[i, j] = i;
}
else if (X[i - 1] == Y[j - 1])
{
dp[i, j] = 1 + dp[i - 1, j - 1];
}
else
{
dp[i, j] = 1 + Math.Min(dp[i - 1, j], dp[i, j - 1]);
}
}
}
// Following code is used to print
// shortest supersequence dp[m,n] s
// tores the length of the shortest
// supersequence of X and Y
int index = dp[m, n];
// string to store the shortest supersequence
String str = "";
// Start from the bottom right corner and one by one
// push characters in output string
i = m; j = n;
while (i > 0 && j > 0)
{
// If current character in X and Y are same, then
// current character is part of shortest supersequence
if (X[i - 1] == Y[j - 1])
{
// Put current character in result
str += (X[i - 1]);
// reduce values of i, j and index
i--;
j--;
index--;
}
// If current character in X and Y are different
else if (dp[i - 1, j] > dp[i, j - 1])
{
// Put current character of Y in result
str += (Y[j - 1]);
// reduce values of j and index
j--;
index--;
}
else
{
// Put current character of X in result
str += (X[i - 1]);
// reduce values of i and index
i--;
index--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str += (X[i - 1]);
i--;
index--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str += (Y[j - 1]);
j--;
index--;
}
// reverse the string and return it
str = reverse(str);
return str;
}
static String reverse(String input)
{
char[] temparray = input.ToCharArray();
int left, right = 0;
right = temparray.Length - 1;
for (left = 0; left < right; left++, right--)
{
// Swap values of left and right
char temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return String.Join("",temparray);
}
// Driver code
public static void Main(String[] args)
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine(printShortestSuperSeq(X, Y));
}
}
/* This code has been contributed
by PrinciRaj1992*/输出：
AGXGTXAYB
上述解决方案的时间复杂度为O(n 2 )。
该程序使用的辅助空间为O(n 2 )。