我读一下CLRS,在练习6.5-8时有一些问题。
Give an O(n lg k)-time algorithm to merge k sorted lists into one sorted list, where n is the total number of elements in all the inputs lists. (Hint: use a min0heap for k-way merging.)
大家都说,解决办法是
(1) 利用k清单的第一要素,建造一个lement粒,
2) 摘录(摘录)以获取最小元素,从肥皂中取到结果清单,
3)从上选取下一个内容,即我们刚刚从中选取的。 插入斜坡,第2段。
我可以理解时间是O(n lg k),但我并没有获得第3步中选择的正确性。 http://www.un.org/?
正确性证据的主旨是,剩下的最低要素始终是要提取的内容。 支持这一索赔的关键是,每个清单的优先顺序都包含该清单中剩下的最低内容。 因此,由于余下的最低部分也是从“<>/em>号”名单上删除的其余最低部分,因此由 extract取回。
我们需要从第3部分的同一清单中选择一个要素,以保持每个清单在调查表中至少有一个要素的不变性。 否则,我们可能会有这样的情况。
1 2
3 4
如果我们从包含1和3的首选中删除1个,将其改为4个,则下一次开采将是3个错误。
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