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I got a naive solution, and I don't know if it's the minimize steps.
The strategy is to loop through the desired order, if desire[i] is empty(-1) or it's just equal to current[i], continue; or else if the current[i] is not empty, we move current[i] to empty slot first, and then move desire[i] to slot i. The following is runnable C++ code:
Cause the algorithm travel the desire oder once, ant the operation of each loop body is constant time, the time complexity is O(n)
- alanluanxl January 01, 2017