打印最短公共超序列
原文:https://www . geesforgeks . org/print-short-common-super quence/
给定两个字符串 X 和 Y，打印同时具有 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，请打印其中任何一个。 例:
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
我们已经讨论了如何打印两个给定字符串的最短可能超序列长度这里。在这篇文章中，我们打印了最短的超序列。 我们已经在下面讨论了在之前的后 中寻找最短超序列长度的算法
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对字符串 X =“AGGTAB”和 Y =“GXTXAYB”、 进行自下而上的求解，则上述算法遵循的步骤
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 <bits/stdc++.h>
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
// 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--;
}
// 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--;
}
else
{
// Put current character of X in result
str.push_back(X[i - 1]);
// reduce values of i and index
i--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str.push_back(X[i - 1]);
i--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str.push_back(Y[j - 1]);
j--;
}
// 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";
cout << printShortestSuperSeq(X, Y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
/* 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
// 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--;
}
// 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--;
}
else
{
// Put current character of X in result
str += (X.charAt(i - 1));
// reduce values of i and index
i--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str += (X.charAt(i - 1));
i--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str += (Y.charAt(j - 1));
j--;
}
// 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 29AjayKumar
Python 3
# 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
# 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
# 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
else:
# Put current character of X in result
string += x[i - 1]
# reduce values of i and index
i -= 1
# If Y reaches its end, put remaining characters
# of X in the result string
while i > 0:
string += x[i - 1]
i -= 1
# If X reaches its end, put remaining characters
# of Y in the result string
while j > 0:
string += y[j - 1]
j -= 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
# sanjeev2552
C
/* 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
// 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--;
}
// 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--;
}
else
{
// Put current character of X in result
str += (X[i - 1]);
// reduce values of i and index
i--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str += (X[i - 1]);
i--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str += (Y[j - 1]);
j--;
}
// 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*/
java 描述语言
<script>
/* A dynamic programming based Javascript program print
shortest supersequence of two strings */
// returns shortest supersequence of X and Y
function printShortestSuperSeq(X,Y)
{
let m = X.length;
let n = Y.length;
// dp[i][j] contains length of
// shortest supersequence
// for X[0..i-1] and Y[0..j-1]
let dp = new Array(m + 1);
for(let i=0;i<(m+1);i++)
{
dp[i]=new Array(n+1);
for(let j=0;j<(n+1);j++)
dp[i][j]=0;
}
// Fill table in bottom up manner
for (let i = 0; i <= m; i++)
{
for (let 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
// string to store the shortest supersequence
let str = "";
// Start from the bottom right corner and one by one
// push characters in output string
let 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--;
}
// 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--;
}
else
{
// Put current character of X in result
str += (X[i-1]);
// reduce values of i and index
i--;
}
}
// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str += (X[i-1]);
i--;
}
// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str += (Y[j-1]);
j--;
}
// reverse the string and return it
str = reverse(str);
return str;
}
function reverse(input)
{
let temparray = input.split("");
let left, right = 0;
right = temparray.length - 1;
for (left = 0; left < right; left++, right--)
{
// Swap values of left and right
let temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return (temparray).join("");
}
// Driver code
let X = "AGGTAB";
let Y = "GXTXAYB";
document.write(printShortestSuperSeq(X, Y));
// This code is contributed by rag2127
</script>
Output
AGXGTXAYB
上述解的时间复杂度为 O(n 2 )。 程序使用的辅助空间为 O(n 2 )。 本文由阿迪蒂亚·戈尔供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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