How would you deal with the previous problem if the palindrome could have an odd or even length?
What will be an ideal response?
If the palindrome has an odd number of characters, the two pointers meet in the middle (i.e., stop when they are the same). If the palindrome has an even number of characters the pointers are adjacent at the end of the search. Therefore we stop on LeftPointer+1 = RightPointer. If we don’t know whether the string to test is odd or even, we have to test both conditions to determine when to stop.
CMP r1,r2 ;test for end (odd case)
BEQ ExitS ;if leftPointer = rightPointer exit success
ADD r1,r5,#1 ;move leftPointer right one place and save in r5
CMP r5,r2 ;if moved leftPointer = rightPointer exit success
BEQ ExitS
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What will be an ideal response?
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