# C/C ++ algorithm learning notes-exchange method

Source: Internet
Author: User

Exchange method:

The procedures of the exchange method are the clearest and simplest. Each time, the current elements are compared and exchanged with the subsequent elements one by one.

[Cpp]
# Include <iostream. h>

Void ExchangeSort (int * pData, int Count)

{

Int iTemp;

For (int I = 0; I <Count-1; I ++)

{

For (int j = I + 1; j <Count; j ++)

{

If (pData [j] <pData [I])

{

ITemp = pData [I];

PData [I] = pData [j];

PData [j] = iTemp;

}

}

}

}

Void main ()

{

Int data [] = {10, 9, 8, 7, 6, 5, 4 };

ExchangeSort (data, 7 );

For (int I = 0; I <7; I ++)

Cout <data [I] <"";

Cout <"/n ";

}

# Include <iostream. h>

Void ExchangeSort (int * pData, int Count)

{

Int iTemp;

For (int I = 0; I <Count-1; I ++)

{

For (int j = I + 1; j <Count; j ++)

{

If (pData [j] <pData [I])

{

ITemp = pData [I];

PData [I] = pData [j];

PData [j] = iTemp;

}

}

}

}

Void main ()

{

Int data [] = {10, 9, 8, 7, 6, 5, 4 };

ExchangeSort (data, 7 );

For (int I = 0; I <7; I ++)

Cout <data [I] <"";

Cout <"/n ";

}

Before comparison | first time | second time | third time | fourth time | fifth time | sixth time

10 9 8 7 6 5 4

9 10 10 10 10 10 10

8 8 9 9 9 9 9

7 7 7 8 8 8 8

6 6 6 6 7 7 7

5 5 5 5 6 6

4 4 4 4 4 5

From the above algorithm, the efficiency is basically the same as that of the bubble method.

Reverse Order (worst case)

First round: 10, 9, 8, 7-> 9, 10, 8, 7-> 8, 10, 9-> 7, 10, 9, 8 (three exchanges)

Round 2: 7, 10, 9-> 7, 9, 10, 8-> 7, 8, 10, 9 (exchange twice)

First round: 7, 8, 10, 9-> 7, 8, 9, 10 (switching once)

Cycles: 6

Number of exchanges: 6

Others:

First round:,->, (exchange once)

Round 2: 7, 10, 8, 9-> 7, 8, 10, 9-> 7, 8, 10, 9 (exchange once)

First round: 7, 8, 10, 9-> 7, 8, 9, 10 (switching once)

Cycles: 6

Number of exchanges: 3

From the running table, the exchange is almost as bad as the bubble. This is true. The number of loops is also 1/2 * (n-1) * n, so the complexity of the algorithm is still O (n * n ). Because we cannot give all the information, we can only tell you that the exchange is equally bad (in some cases, slightly better, in some cases ).

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