"Introduction to Algorithms" Introduction to the third edition of Exercise 2-4 Reverse Pairs _ algorithm

There are many answers to this question on the Internet, such as: Http://blog.csdn.net/qcgrxx/article/details/8005221 said in detail.

Questions about reverse alignment: http://www.cppblog.com/ickchen2/articles/62422.html

But in the last question on how to find the number of reverse pairs in a permutation, I know that many of the online modified merge sort O (NLGN) algorithm can be solved, but I have another use of stack O (n) method, I think it is correct, but by a certain volume is wrong, is very depressing, Below I write down the main process, I hope you can help me.

For example, an introduction to the algorithm, array a <2,3,8,6,1> has 5 reverse pairs, respectively (2,1), (3,1), (8,6), (8,1), (6,1).

So I can do this, using the stack data structure, which holds the number pairs (x,y), where x is the element in array a, and y the number of reverse pairs that can be formed at the beginning of x in the array. The initial stack is empty, the number of reverse pairs is ans=0, and the array is scanned from right to left, as follows:

1. Because the first element is 1 and cannot form a reverse order, the (1,0) is pressed into the stack.

2. Because 6 is larger than the top of the stack, so can be formed (6,1) this reverse, and because 1 corresponding to the reverse order is 0, so 6 corresponding to the reverse is 0++ that 1, so will (6,1) pressure stack, ans++ 1.

3. Because 8 is 6 larger than the top element of the stack, so it can be formed (8,6) This reverse pair, since 8 and 6 can form a reverse pair, then 8 can and with 6 elements are formed in reverse order, that is, 8 can form a 1++=2 reverse pair, so will (8,2) pressure stack, ans+=2 3.

4. Because 3 is smaller than the top element of the stack, the stack is (8,2).

5. Because 3 is smaller than the top element of the stack, the stack is (6,1).

6. Because 3 is larger than the top of the stack, at this time the top of the stack is (x,y) = (1,0), so you can form a y++=1 reverse pair, will (3,1) pressure stack, ans+=1 4.

7. Because 2 is smaller than the top element of the stack, the stack is (3,1).

8. Because 2 is larger than the top element of the stack, it can form a y++=0++=1 reverse pair, 2,1 the stack, ans+=1 5.

9. All elements have been tested, and the final ANS 5 is the number of reverse pairs.

Because the above algorithm simply traverses the array, the time complexity is O (n).

I think the above ideas are reasonable, do not know where there are mistakes, please advise.