对动态规划的引入的问题：Stanford Algorithms: Design and Analysis, Part 2 (week3)

I am very lost in this introduction, and ask for help to get the logic.

Here, the professor introduces four ways:1.greedy 2. divide & conquer 3.linear time 4.reconstruction.

For 1.greedy, it may produce a not maxium result. So it's out.

For 2.divide & conquer, it takes exponential time, too slow. So it's out.

Here, it comes my confusion.

I can't figure out the difference between linear time and reconstruction. Why it costs O(n) and What's the fill-in array used for?

Are both of 3.linear time and 4.reconstruction Dynamical Programming?

感谢大家的帮助！

I can't figure out the difference between linear time and reconstruction. Why it costs O(n)
不知道你上过part I没有，里面讲了时间复杂度的分析吧。dynamic programming和recursion最重要的区别之一就是它复用了以前计算过的subproblem的optimal solution，而不是重复的计算，这就大大节约的时间。这里的cost O(n)只是针对你说的例子，不同情况还得你自己分析。比如到了knapsack problem你会发现其实是pseudo-polynomial。

and What's the fill-in array used for?s
这个table用来reconstruct optimal solution，因为dynamic programming通常得到的是你求的最优的value，而不是解决方法。比如你用dynamic programming 实现Bellman-Ford，得到的只是最短路径的值，但你想知道的是该走哪些edge? 这就需要这个fill-in table来重构了（当然，有时候可以optimal space,不用整个table,这就需要在dynamic programming循环里面做一些标记，记录一些重要信息了）

sprayfly 发表于 2016-8-17 01:16
I can't figure out the difference between linear time and reconstruction. Why it costs O(n)
不知道 ...

非常感谢你！！！
讲的很清晰！