Longest common sub-sequence LCS

Defined:

Subsequence: Part of the original sequence, required to appear in the order of the original sequence, but does not require continuous

Problem Description:

Given two sequences of x=<x1,x2,..., xm> and Y=<y1,y2,..., Yn>, find the maximum length of x and y common sub-sequences

Solving:

Step One: Optimal substructure (describes one of the longest common subsequence LCS)

Definition: xi=<x1,x2,..., xi>,yi=<y1,y2,..., yi>

Assuming that z=<z1,z2,..., zk> as an LCS for X and Y, there are:

1) If Xm=yn, then Zk=xm=yn, and Zk-1 is an LCS of Xm-1 and Yn-1

2) If Xm!=yn, then if ZK!=XM, then Z is an LCS of Xm-1 and yn

If Zk!=yn, then Z is an LCS of X and Yn-1

Step two: Construct recursive solutions

Defined:

C[i,j] Length of an LCS for sequence Xi and Yi

Formula:

C[i,j]=0 if I=0 or J =0

c[i,j]=c[i-1,j-1]+1 if I,j>0 and Xi=yj

C[i,j]=max (C[i-1,j],c[i,j-1]) if i,j>0 and Xi!=yj

Step three: Identify overlapping sub-problems and the number of real sub-problems

Calculate c[i-1,j] and c[i,j-1] will calculate c[i-1,j-1]

As can be seen directly in the form of recursive solutions, the number of sub-problems is θ (MN)

Step Four: Algorithm design

The code is as follows:

intCALCLS (stringS1,stringS2) {    string:: Size_type m=s1.size () +1; string:: Size_type n=s2.size () +1; inti,j; int**c=New int*[M];  for(i=0; i<m;i++) C[i]=New int[n];  for(i=1; i<m;i++) c[i][0]=0;  for(j=0; j<n;j++) c[0][j]=0;  for(i=1; i<m;i++)    {         for(j=1; j<n;j++)            if(s1[i]==S2[j]) c[i][j]=c[i-1][j-1]+1; Else                if(c[i-1][j]>=c[i][j-1]) C[i][j]=c[i-1][j]; ElseC[i][j]=c[i][j-1]; }    returnc[m-1][n-1];}

Longest common sub-sequence LCS