a main difference between set , list set admits no duplicates. instead of list<integer[]> trying create set<integer[]> such no 2 elements equal. getting following results when read set<integer[]>
[0, 4, 5] [3, 4, 1] [4, 5, 0] [0, 3, 6] [1, 3, 4] [1, 2, 7] for implementation, [0, 4, 5] , [4, 5, 0] considered equal. hence question: there way override equals method of integer[] add method of set function can avoid admitting both [0, 4, 5] , [4, 5, 0]?
short answer: can't. there's no mechanism override in array class @ all.
instead, please consider using alternative container objects. use set<integer> objects, creating set<set<integer>>, since seems want order-less comparisons.
or, more fine-tuned control, consider using own class wraps array or set , having e.g. set<myintegerbag>. have full control on comparison operation used.
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