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

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