Comparison of two methods of LCS

The dynamic programming problem generally has two elements: the optimal substructure and the sub-problem overlap.

In general, we use two methods when solving LCS problems:

1. Momo-ization (Memo method)

The overlapping sub-problem characteristics of this problem are exploited, and the overlapping sub-problem can be solved directly by using recursion.

0 A b C b D A b
0 0 0 0 0 0 0 0 0
B 0 0 1 1 1 1 1 1
D 0 0 1 1 1 2 2 2
C 0 0 1 2 2 2 2 2
A 0 1 1 2 2 2 3 2
B 0 1 2 2 3 3 3 4
A 0 1 2 2 3 3 4 4

The so-called top-down is the recursive solution from the largest point of the table below, the final result is LCS (x,y,x.length,y.length);

That is, the above table from the last grid up to the Sentinel process, before solving each sub-problem, we first detect this sub-problem before we have counted, if there is, then do not calculate the direct return results, if not, then calculate this sub-problem, then save the results for the next encounter when the use of.

Time complexity t (n) = O (MN);

Space complexity s = O (MN);//two-dimensional array

[Java] View plaincopy
01.public class Ttblcs {
int[][static] C = new int[100][100];
GENEVA static int NIF = 9999;
public static void Main (string[] args) {
char[] x = {' A ', ' B ', ' C ', ' B ', ' D ', ' A ', ' B '};
Char[] y = {' B ', ' D ', ' C ', ' A ', ' B ', ' A '};
07.
//ttblcs t = new Ttblcs ();
(int i = 0;i <= x.length;i++) {//around a circle of Sentinels are 0
Ten. for (int j = 0;j <= y.length;j++)
11. {
C[I][J] = NIF;
13.}
14.}
System.out.print (LCS (x,y,x.length,y.length));//top-down
16.}
17.
public static int LCS (char[] x,char[] Y,int I,int j) {
if (C[i][j] < NIF)//records are returned directly (memo) if calculated
return C[I][J];
if ((i = = 0) | | (j = = 0)) {
C[I][J] = 0;
23.}
. else if (x[i-1] = = Y[j-1])
C[I][J] = LCS (x,y,i-1,j-1) + 1;
. else
C[I][J] = LCS (X,Y,I-1,J) >= LCS (x,y,i,j-1)? LCS (X,Y,I-1,J): LCS (X,Y,I,J-1);
. return C[I][J];
29.}
30.
31.}

2. Dynamic Programming DP:

The so-called bottom-up, that is, from the subscript (mark) at the beginning of the process of solving, but to eliminate the recursive process, the bottom-up construction of the original problem of the solution, the first to solve the most basic situation, and then from the most basic situation of a part of the upward solution, such as I want to solve [2 ... 4], then I need to know first [2 ... 2][3 ... 4] and [2 ... 3][4 ... 4] The optimal solution, need to know [3 ... 4], then first need to know [3 ... 3][4 ... 4] The optimal solution, so, instead of the original problem needs to be built first, and then slowly up a layer of construction, the final composition of the original problem solution!.

Time complexity t (n) = O (MN);

Space complexity s = O (MN);//two-dimensional array

[Java] View plaincopy
01.public class Bttlcs {
int[][static] C = new int[100][100];
public static void Main (string[] args) {
char[] x = {' A ', ' B ', ' C ', ' B ', ' D ', ' A ', ' B '};
Char[] y = {' B ', ' D ', ' C ', ' A ', ' B ', ' A '};
. for (int k = 0;k <= x.length;k++)
07. {
C[0][k] = 0;
09.}
Ten. for (int k = 0;k <= y.length;k++)
11. {
C[k][0] = 0;
13.}
LCS (x, y);
15.}
16.
. public static void LCS (char[] x,char[] y) {
for (int i = 1;i <= x.length;i++)
19. {
for (int j = 1;j <= y.length;j++)
21. {
if (x[i-1] = = Y[j-1])
C[I][J] = c[i-1][j-1] + 1;
. else
C[I][J] = C[i-1][j] >= c[i][j-1]?c[i-1][j]:c[i][j-1];
26.}
27.}
. for (int i = 0;i <= x.length;i++)
29. {
. for (int j = 0;j<y.length;j++)
31. {
System.out.print (C[i][j]);
33.}
System.out.print ("\ n");
35.}
System.out.print (C[x.length][y.length]);
37.}
38.}

The top-down advantage is that there is no need to solve the solution of each sub-problem, but only the solution of the sub-problem that needs to be solved, the disadvantage is that recursive invocation is required and the function call wastes time.
The bottom-up advantage is that there is no need for recursive calls, each sub-problem solver faster, the shortcomings of each sub-problem to be calculated, even if the solution of this sub-problem to the original problem of the solution does not help!

Comparison of two methods of LCS