# Data structure Learning-array a[m+n] in turn two linear tables (A1,A2 AM), (b1,b2 bn), swap the two sequential table positions

Source: Internet
Author: User

Swaps the two sequential table positions in the array, which will be (b1,b2 bn) put to (A1,a2 AM) front.

Solution One:

Adds all the elements in the array (a1,a2, am,b1,b2, bn) in-situ inversion (bn,bn-1, B1,am,am-1 A1), the first n elements and the rear m elements are respectively reversed, resulting (b1,b2 BN,A1,A2 AM) to achieve position interchangeability.

Code:

`voidReverse (intA[],intLeftintRightintarraySize) {//reversal (aleft,aleft+1,aleft+2,aright) for (aright,aright-1,,aleft)    if(Left>=right| | right>=arrysize)return; intMid= (left+right)/2;  for(intI=0; i<=mid-left;i++)    {        inttemp=a[left+i]; A[left+i]=a[right-i]; A[right-i]=temp; }}        voidExchange (intA[],intMintNintarraySize) {Reverse (A,0, m+n-1, arrysize); Reverse (A,0, N-1, arrysize); Reverse (A,n,m+n-1, arrysize); }`

Solution Two:

Implemented with auxiliary arrays.

To create an array of size m, the first m integers in a are temporarily present in S, while the last n elements in a are shifted to the left, and then the staged in S is placed back in a subsequent cell in a.

``` void  Exchange (int  a[],int  m,int   N) {

int
s[m],i;
for  (I=
0 ; i<m;i++)
//
Save the first m to S  s[i]=
a[i];
for  (I=
0 ; i<n;i++)
//
move back N to front  a[i]=a[m+
i];
for  (I=
0 ; i<m;i++)
//
Insert the scratch in S in the back  a[n+i]=
s[ I];}
```

Data structure Learning-array a[m+n] in turn two linear tables (A1,A2 AM), (b1,b2 bn), swap the two sequential table positions

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