Finding ONE good VLSI chip in a population of good and bad ones, by using the
pair test.
Chip A Chip B Conclusion
------- ------- ----------
B is good A is good both are good or both are bad
B is good A is bad at least one is bad
B is bad A is good at least one is bad
B is bad A is bad at least one is bad
Assumption : number of goods > number of bads
We can solve this in O(n) time complexity by splitting the population in half
every time and collecting one element of the GOOD, GOOD pair.
T(n) = T(n/2) + n/2
But to collect the pairs we need n/2 memory separately.
Can we do this in-place without using extra memory ??
答案 0 :(得分:0)
该算法基于以下问题:“我们可以删除这个芯片吗?”因此,对于要删除的每个芯片,我们只需将其从链接列表中删除(或者更确切地说,根本没有位置)。