Determining a sequence is sorted or not

If you have a list or array of elements then how should we determine the given sequence is either sorted in ascending or descending order.

There are many ways of finding it  but the following solution provides a solution in one iteration so complexity is O(n).

We need to compare the elements such that either elements are either increasing or decreasing  , if this condition is not matching then unsorted sequence otherwise its a sorted sequence.

/**

 * @author sachin.srivastava

 * @tek9g.blogpspot.com

 * {@code checking various scenarios of handling the sorting cases.}

 */

//checking if a list is sorted means asceding or descending in either way.

//complexity is O(n) beacaue of traversal of n elements of the list

public static <T extends Comparable<T>> boolean isSorted(List<T> list) {

if(list == null || list.isEmpty()){

return false;

}

if(list.size() == 1) {

return true;

}

T prev =null;

boolean asc = true;

boolean desc = true;

for (T t : list) {

//ascending

if(prev!=null && prev.compareTo(t) > 0 ){

asc = false;

} 

//descending

if(prev!=null && prev.compareTo(t) < 0 ){

desc = false;

}

prev  = t;

}

return asc || desc;

}

Testcases : 

            System.out.println(isSorted(Arrays.asList(1,2,4,5)));//true

System.out.println(isSorted(Arrays.asList(5,2,4,5)));//false

System.out.println(isSorted(Arrays.asList("abc","bcd","def","fgh")));//true

System.out.println(isSorted(Arrays.asList("b","a","c","d")));//false

Enjoy the coding !!!!