Common sorting by C language

# Include <stdio. h>
# Include <conio. h>
# Define N 10
Int main (void)
{
Int I, j, temp;
Int A [n] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

// Sort by Bubble Sorting
For (I = 0; I <n-1; I ++ ){
For (j = 0; j <N-i-1; j ++ ){
If (A [I] <A [J]) {
Temp = A [I];
A [I] = A [J];
A [J] = temp;
}
}
} // The first

Printf ("the ten numbers you enter are:/N ")
;
For (I = 0; I <n; I ++ ){
Printf ("% d", a [I]);
}
Getch ();
Return 0;
}

 

The following describes how to select a sort method, which is more efficient than the bubble sort method:

# Include <stdio. h>
# Include <conio. h>
# Define N 10
Int main (void)
{
Int I, J, K, temp;
Int A [n] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

// Select sorting method for sorting
For (I = 0; I <n-1; I ++ ){
K = I;
For (j = I + 1; j <n; j ++ ){
If (A [k] <A [J]) {
K = J;
}
Temp = A [I];
A [I] = A [k];
A [k] = temp;
}
} // The first

Printf ("the ten numbers you enter are:/N ")
;
For (I = 0; I <n; I ++ ){
Printf ("% d", a [I]);
}
Getch ();
Return 0;
}

 

The insert sorting method is more efficient than the Bubble sorting method and the select sorting method:

# Include <stdio. h>
# Include <conio. h>
# Define N 10
Int main (void)
{
Int I, j, key, temp;
Int A [n] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

// Sort by insert sorting
For (I = 1; I <n; I ++ ){
Key = A [I];
For (j = I-1; j> = 0; j --){
If (A [J] <key ){
A [J + 1] = A [J];
}
}
A [J + 1] = key;
} // The first

Printf ("the ten numbers you enter are:/N ")
;
For (I = 0; I <n; I ++ ){
Printf ("% d", a [I]);
}
Getch ();
Return 0;
}

The following is the Merge Sorting method. I wrote it after reading introduction to algorithms. I cannot write it myself. Code writing is much easier than writing in other books. The efficiency of the Merge Sorting method is that these four sorting methods
The most efficient:

# Include <stdio. h>
# Include <conio. h>
# Define N 10

Void merge_sort (int *,
Int P, int Q, int R)
{
Int I, J, K, M, N;
Int array1 [N], array2 [N];

// M and n are two newly created arrays respectively.
// Number of elements
M = Q-p + 1;
N = r-Q;

// First put the elements in array a [n]
// In the newly created two Arrays
For (I = 0; I <m; I ++ ){
Array1 [I] = * (a + P + I );
}
For (j = 0; j <n; j ++ ){
Array2 [J] = * (a + q + J + 1 );
}

I = J = 0;
K = P; // K cannot be initialized to 0

// Start merging
While (I <M & J <n ){
If (array1 [I]> array2 [J]) {
* (A + k) = array1 [I];
++ K;
++ I;
}
Else {
* (A + k) = array2 [J];
++ K;
++ J;
}
}

// Merge the remaining elements of each array
While (I <m ){
* (A + k) = array1 [I];
++ K;
++ I;
}
While (j <n ){
* (A + k) = array2 [J];
++ K;
++ J;
}
}

Void Merge (int * a, int P,
Int R)
{
Int Q;

// If P> = r, there is only one element to be sorted,
// That is, it has been sorted
If (P <r ){
Q = (p + r)/2;
Merge (A, p, q );
Merge (A, q + 1, R );
Merge_sort (A, P, Q, R );
}
} // Merge

Int main (void)
{
Int I, j, key, temp;
Int A [n] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

Merge (A, 0, 9 );
Printf ("the ten numbers you enter are:/N ");
For (I = 0; I <10; I ++ ){
Printf ("% d", a [I]);
}
Getch ();
Return 0;
}

 

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