n elements into the stack, how many kinds of stack order?

Original: http://blog.csdn.net/zyearn/article/details/7758716

Recently in reviewing data structure, see the stack, found 1 elements into the stack, there are 1 kinds of stack order, 2 elements into the stack, 2 kinds of stack order, 3 elements into the stack, there are 5 kinds of stack order, then a very natural problem is n elements into the stack, how many kinds of stack order?

I'm ashamed to say that I didn't think about it when I learned the data structure. Recently in the dynamic planning, so the "sub-problem" These 3 words have been wandering in my mind, so when solving this problem I also use a similar "sub-problem" method, plainly speaking is the recursive formula.

We put the number of n elements out of the stack as f (n), then for the three-to-one, we can easily conclude:

F (1) = 1//or 1

F (2) = 2//i.e. 12, 21

F (3) = 5//i.e. 123, 132, 213, 321, 231

Then we consider F (4), we give 4 elements to a,b,c,d, then consider: element A can only appear in position 1th, position 2nd, position 3rd and position 4th (very easy to understand, altogether 4 locations, such as ABCD, element A is in position 1th).

Analysis:

1) If element A is in position 1th, then only a stack is possible, immediately out of the stack, then there are elements B, C, d wait for operation, is the sub-problem F (3);

2) If element A is in position 2nd, then there must be an element in the first stack than a, that is, f (1) possible order (only B), and the remaining C, D, that is, F (2), according to the multiplication principle, the total number of orders is F (1) * f (2);

3) If element A is in position 3rd, then there must be two elements in the stack than 1, that is, F (2) possible order (only B, C), and D, that is f (1),

According to the multiplication principle, the total order number is F (2) * f (1);

4) if element A in position 4th, then must be a advanced stack, and finally out of the stack, then the order of the elements B, C, D is the solution of this small problem, that is, F (3);

Combine all cases, i.e. f (4) = f (3) + F (2) * F (1) + f (1) * F (2) + f (3);

To be normalized, we define f (0) = 1, and f (4) can be re-written as:

F (4) = f (0) *f (3) + F (1) *f (2) + F (2) * F (1) + F (3) *f (0)

Then we generalize to N, the idea of promotion and n=4 is exactly the same, so we can get:

F (n) = f (0) *f (n-1) + f (1) *f (n-2) + ... + f (n-1) *f (0)

That

But this is just a recursive formula (if implemented programmatically, a one-dimensional array needs to be maintained, with a time complexity of O (n^2)).  How do you turn it into a general formula, with only an O (1) complexity? So search on the internet, there really is such a formula: C (2n,n)/(N+1) (c (2n,n) for 2n), and a name called Catalan number. Attached to the wiki link, written in too detail: Http://en.wikipedia.org/wiki/Catalan_number Now the question is: How to find out from the above recursive formula C (2n,n)/(n+1)? Interested friends welcome message to discuss! 2013.6.4 Update According to the idea of netizens u010896627, I abstract the next question, in the knowledge asked a question, one of the answers to the "discounted method", from the geometry of the "n elements into the stack number of stack order" the answer to this question is C (2n,n)-B ( 2n,n-1), Catalan number is a simple simplification. We recommend that you take a look.

n elements into the stack, how many kinds of stack order?