Dynamic programming---The longest common subsequence

1, problem description
A subsequence of a given sequence is a sequence that is obtained after deleting several elements in the sequence. Specifically, if the given sequence x={x1,x2,..., XM}, then another sequence of Z={Z1,Z2,..., the zk},x sub-sequence refers to the existence of a strictly incrementing subscript sequence {i1,i2,..., ik} so that for all j=1,2,... K has zj=xij for example, sequence Z={b,c,d , B} is a subsequence of sequence x={a,b,c,b,d,a,b}, and the corresponding increment subscript sequence is {2,3,5,7}.
Given two sequences x and Y, when another sequence z is both a subsequence of X and a sub-sequence of y, it is said that Z is the common subsequence of the sequence x and Y, that the common subsequence can have multiple, and that the longest common subsequence is the longest of the x and Y sequences.
2. Algorithm Description:
When XM =yn, find the longest common subsequence of Xm-1 and Yn-1, and then add XM (=yn) at the end of it to get the longest common subsequence of x and Y. When XM!=yn, you must solve two sub-problems, that is, to find a longest common subsequence of Xm-1 and Y and X and Yn-1 One of the longest common subsequence sequences. The older of the two common subsequence sequences is the longest common subsequence of x and Y.

C[I][J] The length of the longest common subsequence storing XI and YJ, From[i][j] representing the length of the c[i][j] is derived from which case;

3. Algorithm time complexity analysis
The length complexity of the common sub-sequence is m*n, the sub-sequence is obtained according to the Storage state table and the output complexity is m*n, so the final time complexity is m*n.

The code is as follows:

ImportJava.util.Scanner; Public classLCS { Public Static voidMain (string[] args) {//Abcdefgie//BCDFGEScanner input =NewScanner (system.in); String as=Input.nextline (); String BS=Input.nextline (); Char[] A =As.tochararray (); Char[] B =Bs.tochararray (); int[] from =Init (A, b);        Comlength (A,b,from); //PrintArray (from);Printcom (a,from,a.length-1,b.length-1); }     Public Static voidPrintArray (int[] from) {//enter the given array to print it         for(inti=0; I<from.length; i++){             for(intj=0; J<from[0].length; J + +) {System.out.print (From[i][j]+" ");        } System.out.println (); }    }     Public Static int[] Init (Char[]a,Char[]b) {//creating an array of sources        int[] from =New int[A.length][b.length];  for(inti=0; i<a.length; i++){             for(intj=0; j<b.length; J + +) {From[i][j]=0; }        }        returnFrom ; }     Public Static voidComlength (Char[]a,Char[]b,int[][]from) {//assigning values to the from state array         intm =a.length; intn =b.length; int[][]C =New int[M][n];  for(inti=0; i<m; i++){              for(intj=0; j<n; J + +) {C[i][j]=0; }         }          for(intI=1; i<m; i++){              for(intJ=1; j<n; J + +){                 if(A[i]==b[j]) {//The representative derives from the 1th case: The last one of the two strings of the comparison is equalC[i][j]=c[i-1][j-1]+1; FROM[I][J]=1; }Else if(C[i][j-1]>=c[i-1][j]) {//The representative derives from the 2nd case: The last one of the last equal and longest common subsequence of the two strings compared is not equal to the last one of the array of a character sequences;C[i][j]=c[i][j-1]; FROM[I][J]=2; }Else{//The representative derives from the 3rd case: The last of the last unequal and longest common subsequence of the two strings compared is not equal to the last one of the B array character sequences;C[i][j]=c[i-1][j]; FROM[I][J]=3; }             }         }//PrintArray (c);//System.out.println ();    }     Public Static voidPrintcom (Char[]a,int[][]from,intMintN) {//outputs the longest common subsequence based on the from state array          if(m==0|n==0)return; if(from[m][n]==1) {printcom (a,from,m-1,n-1);          System.out.print (A[m]); }Else if(from[m][n]==2) {printcom (A,from,m,n-1); }Else{printcom (a,from,m-1, N); }    }}

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Dynamic programming---The longest common subsequence