Longest common sub-sequence-dynamic planning

A sub-sequence of a string is a sequence of elements that are removed from the sequence, such as a helloworld of a hold.

Consider how the longest common subsequence problem is decomposed into sub-problems, set a= "A0,a1,...,am-1", b= "B0,b1,...,bm-1", and z= "Z0,z1,...,zk-1" as their longest common subsequence.

It is not difficult to prove the following properties:

(1) If am-1=bn-1, then zk-1=am-1=bn-1, and "Z0,z1,...,zk-2" is a "a0,a1,...,am-2" and "b0,b1,...,bn-2" a longest common sub-sequence;

(2) If the am-1!=bn-1, if the zk-1!=am-1, the implication "z0,z1,...,zk-1" is "a0,a1,...,am-2" and "b0,b1,...,bn-1" a longest common sub-sequence;

(3) If am-1!=bn-1, then if zk-1!=bn-1, implication "z0,z1,...,zk-1" is a "a0,a1,...,am-1" and "b0,b1,...,bn-2" a longest common sub-sequence.

Thus, when looking for the common sub-sequences of A and B, if there is a am-1=bn-1, then further solve a sub-problem, find "a0,a1,...,am-2" and "b0,b1,...,bm-2" a longest common sub-sequence;

If am-1!=bn-1, solve two sub-problems, find one of the longest common subsequence of "a0,a1,...,am-2" and "b0,b1,...,bn-1" and find a longest common subsequence of "a0,a1,...,am-1" and "B0,b1,...,bn-2", Then take the older elders as the longest common subsequence of a and B.

Define LCS[I][J] for the length of the longest common subsequence of the sequence "a0,a1,...,ai-1" and "b0,b1,...,bj-1", the calculation Lcs[i][j] can be recursively expressed as follows:

(1) Lcs[i][j] = 0 if i=0 or j=0;

(2) Lcs[i][j] = lcs[i-1][j-1]+1 if i,j>0, and a[i-1] = b[j-1];

(3) Lcs[i][j] = Max{lcs[i][j-1], Lcs[i-1][j]} if i,j>0, and a[i-1]! = b[j-1].

By this calculation, you can write out the length function that calculates the longest common subsequence of two series. Because the generation of lcs[i][j] depends only on lcs[i-1][j-1], lcs[i-1][j] and lcs[i][j-1], it is possible to start with lcs[m][n], and then construct the longest common sub-sequence in reverse. See the procedure for details.

1  PackageLCS;2 3  Public classLCS {4     Private Static voidcomsubstr (String str1,string str2) {5         Char[] A=Str1.tochararray ();6         Char[] b=Str2.tochararray ();7         intalen=a.length;8         intblen=b.length;9         int[] lcs=New int[Alen+1] [Blen+1];Ten          One          for(inti=0;i<=alen;i++) A              for(intj=0;j<=blen;j++) -Lcs[i][j]=0; -          the          for(inti=1;i<=alen;i++) -              for(intj=1;j<=blen;j++) -             { -                 if(a[i-1]==b[j-1]) +Lcs[i][j]=lcs[i-1][j-1]+1; -                 Else { +Lcs[i][j]=lcs[i][j-1]>lcs[i-1][j]?lcs[i][j-1]:lcs[i-1][j]; A                 } at             } -          -          for(inti=0;i<=alen;i++) -         { -              for(intj=0;j<=blen;j++) -             { inSystem.out.print (lcs[i][j]+ ","); -             } to System.out.println (); +         } -          the         intmaxlen=Lcs[alen][blen]; *         intI=ALen; $         intj=Blen;Panax Notoginseng         Char[] comsubstr=New Char[MaxLen]; -          while(maxlen>0) the         { +             if(Lcs[i][j]!=lcs[i-1][j-1])//stating that ai-1 or bj-1 are public in a0......ai-1,b0.......bj-1; A             { the                 if(Lcs[i][j-1]==lcs[i-1][j])//Description Ai-1=bj-1 is a public character; +                 { -Comsubstr[maxlen-1]=a[i-1]; $maxlen--; $i--; -j--; -                 } the                 Else{//take the longer of the two common sub-sequences as a, B, -                     if(lcs[i][j-1]>lcs[i-1][j])Wuyij--; the                     Else { -i--; Wu                     } -                 } About             } $             Else{ -i--; -j--; -             } A         } +System.out.print ("The longest common string is:"); the System.out.println (COMSUBSTR); -     } $      Public Static voidMain (string[] args) { theString a = "blog.csdn.net";  theString B = "Csdn.blogt"; the Lcs.comsubstr (A, b); the     } -}

The output is:

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,2,2,2,2,0,0,0,0,0,0,1,2,3,3,3,0,0,0,0,0,0,1,2,3,4,4,0,0,0,0,0,1 , 1,2,3,4,4,0,1,1,1,1,1,1,2,3,4,4,0,1,2,2,2,2,2,2,3,4,4,0,1,2,3,3,3,3,3,3,4,4,0,1,2,3,4,4,4,4,4,4,4,0,1,2,3,4,5,5,5,5,5,5 , 0,1,2,3,4,5,5,5,5,5,5,0,1,2,3,4,5,5,5,5,5,5,0,1,2,3,4,5,5,5,5,5,6, the longest common string is: csdn.t

　　

Longest common sub-sequence-dynamic planning