REAMS.AL
BAN USER
Comments (9)
Reputation 10
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of 1 vote
Take nod->data from all the nodes and concatenate in string1, then take two pointer, one
moving forward another backward ,move both pointers one by one from both sides check to see if characters are equal till pointers cross esch other, return true if all charaters match else false...........
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see inplace means you cannot use any other data structure , not even another list, and you have to do this in o(n) time , if you traverse the list again it is ok...but you cannot use any other ds also....now how would you do in o(n) time?
- REAMS.AL October 25, 2012