The complexity of the algorithm is seen by a small program

was asked to write a procedure today, the topic is as follows; the price of gold in the last four weeks within 20 trading days, the maximum difference in the last 20 days?

A[20]={2,5,1,6,7,9,10,18...1}

What I was thinking was the difference between the two days of trading, the difference between the 1th and 2nd days, the difference between 2nd and 3rd days, the difference between the first day and the third day, the maximum between them??

So, confidently wrote:

    int a[20]={2,5,1,6,7,9,10,18...1};    int b[19]={0};    int max=0;    for (int m=1;m<20;m++)    {int tempmax=0;//stores the difference Value data for (int n=0;n<19;n++) {b[n]=0;} Calculates the difference for (int i=m;i<m;m++) {b[i]=a[i]-a[i-1];} The maximum value for the difference value for (int n=0;n<19;n++) {if (B[n]>tempmax) {              tempmax=b[n];}} if (Tempmax>max) {Max=tempmax;}    }

Two days of difference, my algorithm complexity n (n-1) * (n-2) **1;

Students, do I write the code right? How well the code is written!! I'm proud of it.

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I want to be any two days between the maximum difference, not necessarily two days between the difference, we do not two days between the difference, turn this bend over, good to do.

I get the maximum value and the minimum value, then the difference between them is not the maximum difference between any two days??

1 inta[ -]={2,5,1,6,7,9,Ten, -...1};2     //Maximum Value3     inttempmax=a[0];4     //Minimum Value5     inttempmin=a[0];6     intMax;7      for(intm=1;m< -; m++)8     {9         if(a[m]>Tempmax)Ten         { Onetempmax=A[m]; A         } -         Else if(a[m]<tempmin) -         { thetempmin=A[m]; -         } -     } -     //as a result, even if we need +Max= tempmax-tempmin;

The complexity of the algorithm is n.
Different ways of implementation, the gap is very large ...

Oh, think more, the way is always more difficult!

The complexity of the algorithm is seen by a small program