Longest [increment] subsequence: note that there is no public sequence, that is, there is only one sequence.

Source: Internet
Author: User

Monotonic: monotonous.

The sequence is X = (x1, x2, X3, x4...). First, sort X to get x', and find the longest common subsequence (LCS) of x and x.



Another way of thinking:

First, review the classical O (N ^ 2) dynamic planning algorithm, set a [I] to representNumber,F [I] indicates the length of the longest ascending subsequence ending with I from 1 to I.In the initial phase, F [I] = 0; (I = 1, 2,..., Len ()). Then there is a dynamic planning equation: F [I] = max {1, F [J] + 1} (j = 1, 2 ,..., i-1, and a [J] <A [I]).


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