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I think this question can be solved by using 2 priority queues.
- Aiolos1127 April 03, 2010Create data structure Node(VAL, INDEX). Index is the index value in the original array, VAL is its value for that index.BTW, the first priority queue compares elements by the its VAL value. And the second priority queue compares each other by its index value.
Step 1: Put all the nodes into the 1st priority queue.
Step 2: keep poping the node from 1st priority queue and push this node into 2nd priority queue if their value is consecutive. If not consecutive, pop and print all the elements in the 2nd priority queue.
Step 3: continue step 2 until the 1st priority queue is empty.