Longest common subsequence and longest increment sub-sequence

1. The longest common subsequence: (x and Y are the lengths of two arrays)

f (x, Y) = 0 if (x==0 | | y==0)

F (x-1,y-1) +1 if (a[x-1]==b[y-1])

Max{f (X-1,y), F (x,y-1)} if (A[x-1]!=b[y-1])

2, the longest increment sub-sequence

(1) The longest common subsequence method: the longest common subsequence of the original array after sorting.

(2) Dynamic programming Method: (Time complexity O (n^2))

An array of length n is {a0,a1, A2, ... an-1), then L (j) ={max{1,l (i) +1}, I<j and A[i]<a[j]} are assumed to be the longest increment subsequence length of an array sequence ending in AJ. That is, we need to traverse all positions before J (from 0 to j-1), find the L (i) that satisfies the condition a[i]<a[j], and calculate the value of Max (L (i)) +1, which is L (j). Finally, we traverse all the L (j) (from 0 to N-1) and find out the maximum is the maximum increment subsequence. The time complexity is O (n^2).
For example, the given array is {5,6,7,1,2,8}, then L (0) =1, L (1) =2, L (2) =3, L (3) =1, L (4) =2, L (5) = 4. So the longest increment of the array is the length of the subsequence 4, and the sequence is {5,6,7,8}. The algorithm code is as follows:

int lis (int arr[], int len)
{    intLongest[len];
 for(intj =1; J < Len; J + +)    {
Longest[i] = 1;  for(inti =0; I < J; i++)        {            if(Arr[j] > Arr[i] && longest[j] < Longest[i] +1) {Longest[j]= Longest[i] +1; }        }    }    intMax =0;  for(inti =0; i < Len; i++)    {        if(Longest[i] >max) Max=Longest[i]; }    returnMax;}

(3)

Longest common subsequence and longest increment sub-sequence