Array Loop Movement

Array Loop Movement

Personal information: studied at Yan University undergraduate software engineering major at present;

I blog: Google search "cqs_2012" can be;

Personal Hobbies: Love data structure and algorithms, hope to work in the future algorithm to make their own contribution to the people;

Programming Language: C + +;

Programming Bad Environment: Windows 7 Professional Edition x64;

Programming Tools: VS2008;

Cartographic tools: Office in PowerPoint;

Hardware information: 7g-3 notebook;

Mantra

Go ahead, don't hesitate, hesitate is the best chance for others to surpass themselves.

Topic

The array moves right through.

Solution

I designed the algorithm time complexity of O (n), the spatial complexity of O (1)

The algorithm is as follows: N is an array length, K is the number of moving digits (K great article here)

First consider if k<0, that is to move to the left;

In consideration if k>n, this is to explain that K has moved several times, it is not necessary, if k = = N, that is, after the move, the array has not changed; what do we do then take k = k%n;

My algorithm is divided into two cases.

d = min (k, n-k),

The first case is n = = = 0 (this needs to find an example, try more, and then you can find the pattern)

The second case is n% d! = 0 (This is a relatively simple case)

C + + code representation algorithm

Array loop shift void Array::roat_right_array (int *data,unsigned int const length,unsigned int k) {//Exception input if (data = NULL | | length = = 0) {cout<< "Abnormal input Roat_right_array" <<ENDL;} Normal input else{//If you move to the left, turn to the right while (K < 0) K + = Length;k = k% Length;//If the entire array does not change after the move, then you do not need to move, otherwise you need to move, as follows if (k! = 0) {int min = (k < (LENGTH/2))? k:length-k;//can divide evenly, meaning is not a dragon if (length% min = = 0) {int dlater,dbefore;for (int j=0;j<min;j++) {dlater = data[(j+k)%length ];d before =data [j];for (unsigned int m=0,i=j;m < (length/min); m++) {data[(i+k)%length] = Dbefore;i = (i+k)%length; Dbefore = Dlater;dlater = data[(i+k)%length];}}} Not divisible, meaning is dragon else{int dbefore= data[0],dlater = data[k%length];for (unsigned int i=0,j=0;i<length;i++) {data[(j+k)% Length] = Dbefore;j = (j+k)%length;dbefore = dlater;d later = data[(j+k)%length];}}}}