问题
I have understood that to find the solution of LIS problem, we need to find a LIS for every subsequence starting from initial element of the array to the each element that ends with a particular element(the last element), but I am not able to understand how would that help in finally finding a LIS of a given unsorted array, I also understand that this leads to an optimal substructure property and then can be solved, but as mentioned, I dont see how finding LIS(j) that ends with arr[j] will help us.
thanks.
回答1:
Consider this sequence as an example:
a[] : 10 20 1 2 5 30 6 8 50 5 7
It produces the following sequence of LIS[i]:
a[] : 10 20 1 2 5 30 6 8 50 5 7
LIS[] : 1 2 1 2 3 4 4 5 6 3 4
Given this sequence, you can immediately find the length of the result, and its last element: the length is 6, and the last element is 50.
Now you can unfold the rest of the sequence, starting from the back: looking for LIS of 5 (one less than that of element 50) such that the number is less than 50 yields 8. Looking back further for 4 gives you 6 (there is no tie, because 30 is above 8). Next comes 5 with LIS of 3, and then a 2 with LIS of 2. Note that there is no tie again, even though 20 has the same LIS. This is because 20 is above 5. Finally, we find 1 with LIS of 1, completing the sequence:
50 8 6 5 2 1
Reversing this produces the longest increasing subsequence:
1 2 5 6 8 50
This is a common trick: given a table with the value of the function that you are maximizing (i.e. the length) you can produce the answer that yields this function (i.e. the sequence itself) by back-tracking the steps of the algorithm to the initial element.
来源：https://stackoverflow.com/questions/43236912/how-does-finding-a-longest-increasing-subsequence-that-ends-with-a-particular-el