C ++ data structure and algorithm _ 2 _ linear table, data structure and algorithm _ 2

C ++ data structure and algorithm _ 2 _ linear table, data structure and algorithm _ 2
Application Example of sequence table-parallel operation and intersection operation of sequence table

# Include "SeqList. h "using namespace std; // calculate void unionSL (SeqList <int> & LA, SeqList <int> & LB) {int m = LA. length (); int n = LB. length (); for (int I = 1; I <= n; ++ I) {int ival; LB. getData (I, ival); // retrieves int searchVal = LA from the LB table. search (ival); if (searchVal = 0) // if the element does not exist in the LA table {LA. insert (m, ival); // Insert to + + m ;}}// submit the void intersectionSL (SeqList <int> & LA, SeqList <int> & LB) {int m = LA. length (); in T I = 1; while (I <= m) {int ival; LA. getData (I, ival); // retrieves int searchVal = LB from LA. search (ival); if (searchVal = 0) // if the data is not found in LB {LA. remove (I, ival); // delete it -- m; // shorten the table length [do not forget!]} Else {++ I ;}}// test program int main () {freopen ("input", "r", stdin); SeqList <int> L1, L2; l1.input (); L2.input (); unionSL (L1, L2); L1.output (); cout <endl; SeqList <int> L3; L3.input (); intersectionSL (L3, l2); L3.output ();}

/** Test data ** 5 1 2 3 4 5 6*4 11 2 3 4 9 ** 4 11 12 13 4 19 */

/* Output result (because redirection is used, the input data is not displayed on the screen )*/

Data Structure algorithm 21 two linear tables la and lb, the new set that, how to write the C language version of the complete program

I wrote it. compile it and change it.

Void Union (LinkList * L1, LinkList * L2, LinkList * & L3) // Intersection
{
LinkList * p = L1-> next, * q = L2-> next, * s, * c;
L3 = (LinkList *) malloc (sizeof (LinkList ));
L3-> next = NULL;
C = L3;
While (p! = NULL & q! = NULL)
{If (p-> data <q-> data)
{S = (LinkList *) malloc (sizeof (LinkList); // copy Node
S-> data = p-> data;
C-> next = s; c = s;
P = p-> next;
}
Else if (p-> data> q-> data)
{S = (LinkList *) malloc (sizeof (LinkList ));
S-> data = q-> data;
C-> next = s; c = s;
Q = q-> next;
}
Else
{
S = (LinkList *) malloc (sizeof (LinkList ));
S-> data = p-> data;
C-> next = s; c = s;
P = p-> next;
Q = q-> next;
}
}

While (q! = NULL)
{
S = (LinkList *) malloc (sizeof (LinkList ));
S-> data = q-> data;
C-> next = s; c = s;
Q = q-> next;
}
C-> next = NULL;

While (p! = NULL)
{
S = (LinkList *) malloc (sizeof (LinkList ));
S-> data = p-> data;
C-> next = s; c = s;
P = p-> next;
}
C-> next = NULL;

}


Intersection of two linear tables A and B in the data structure Algorithm

Sort A and B respectively, and then submit.
For example, sort A and B in ascending order, Set A to P, and B to Q. If A [P]> B [Q], Q ++, if A [P] <B [Q], then P ++; if A [P] = B [Q], Q ++, P ++, Count ++, and [Count] = B [Q-1]; when one of P or Q reaches the end of A or B, the algorithm ends.

The following is a reference program:
//----------------------------------------

# Include <stdio. h>

Int A [100001], B [100001];
Int Ans [100001], Count, N, M;

Void Swap (int & a, int & B)
{
Int Temp =;
A = B;
B = Temp;
}

Void Sort (int L, int R, int A [])
{
Int p, q, Mid;
If (L = R) return;
Mid = A [(L + R)/2];
P = L-1; q = R + 1;
Do
{
P ++; q --;
While (A [p] <Mid) p ++;
While (A [q]> Mid) q --;
If (p <q) Swap (A [p], A [q]);
} While (p <q );
Sort (L, q, );
Sort (q + 1, R, );
}

Void Init ()
{
Int p, q;
P = q = 1;
For (int I = 2; I <= N; I ++)
{
If (A [I]! = A [p])
{
P ++;
A [p] = A [I];
}
}
For (int I = 2; I <= M; I ++)
{
If (A [I]! = B [q])
{
Q ++;
B [q] = A [I];
}
}

}

Int main (void)
{
Int p, q;
Scanf ("% d", & N, & M); // enter the number of elements in the Two Sets
For (int I = 1; I <= N; I ++)
Scanf ("% d", & A [I]); // read A set
For (int I = 1; I <= M; I ++)
Scanf ("% d", & B [I]); // read the B set
Sort (1, N, A); // Sort A set
Sort (1, M, B); // Sort the Set B
Init (); // removes the same elements from the same set
P = q = 1;
Count = 0;
While (p <= N & q <= M) // solves the problem.
{
If (A [p] <B [q])
{
P ++;
Continue;
}
If (A [p]> B [q])
{
Q ++;
Continue;
}
If (A [p] = B [q])
{
P ++; q ++;
Count ++;
Ans [Count] = B [q-1];
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