[置顶] 最长递增子序列

（1）动态规划O（N^2）复杂度

（2）O（N Log N）优化算法

参见文章： http://www.geeksforgeeks.org/longest-monotonically-increasing-subsequence-size-n-log-n/

From the observations, we need to maintain lists of increasing sequences.

In general, we have set of active lists of varying length. We are adding an element A[i] to these lists. We scan the lists (for end elements) in decreasing order of their length. We will verify the end elements of all the lists to find a list whose end element is smaller than A[i] (floor value).

Our strategy determined by the following conditions,

1. If A[i] is smallest among all end candidates of active lists, we will start new active list of length 1.

2. If A[i] is largest among all end candidates of active lists, we will clone the largest active list, and extend it by A[i].

3. If A[i] is in between, we will find a list with largest end element that is smaller than A[i]. Clone and extend this list by A[i]. We will discard all other lists of same length as that of this modified list.

Note that at any instance during our construction of active lists, the following condition is maintained.

“end element of smaller list is smaller than end elements of larger lists”.

It will be clear with an example, let us take example from wiki {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}.