2.2-1:

 O(n^3)

2.2-2:

Selection-Sort(A):

　　for( i=0 ; i<A.size()-1 ; i++){

　　　　max = A[i]; index = 0;

　　　　for(j=i+1; j<A.size(); j++){

　　　　　　if(max < A[j]){

　　　　　　　　max = A[j];

　　　　　　　　index = j;

　　　　　　}

　　　　}

 

　　　　// then swap the next max and the current element

　　　　A[j] = A[i];

　　　　A[i] = max;

　　}

2.2-3:

Say we're sorting n elements

Worst Case: n

Best Case: 1

Average: (1 + 2 + … + n)/n = (n+1)/2

2.2-4:

Know exactly what's the best case, then print it.

(I remember there was an UVA problem simply asking for an answer)