最長遞增子序列 子串

Description:

描述：

This is one of the most popular dynamic programming problems often used as building block to solve other problems.

這是最流行的動態編程問題之一，通常用作解決其他問題的基礎。

Problem statement:

問題陳述：

Given a sequence A of size N, find the length of the longest increasing subsequence from the given sequence.

給定一個大小為N的序列A ，從給定序列中找到最長的遞增子序列的長度 。

The longest increasing subsequence means to find a subsequence of a given sequence where the subsequence's elements are sorted in increasing order, and the subsequence is longest possible. This subsequence is not necessarily contiguous, or unique. Longest increasing subsequence is strictly increasing.

最長的增長子序列意味著找到給定序列的子序列，其中該子序列的元素按升序排序，并且該子序列可能最長。 該子序列不一定是連續的或唯一的。 最長增加的子序列嚴格增加。

Input:N=7Sequence:{2, 3, 4, 0, 1, 2, 3, 8, 6, 4}Output:Length of Longest increasing subsequence is 5Longest increasing subsequence= {0, 1, 2, 3, 8} or {0, 1, 2, 3, 4}

Explanation with example

舉例說明

The possible increasing sub-sequences are,

可能增加的子序列是

Of Length 1 //each element itself is an increasing sequence

長度為1 //每個元素本身都是遞增序列

So, on...

所以...

So, on...

所以...

So, on...

所以...

No more
Of Length 6
None

不再
長度6
沒有

So, the longest increasing subsequence length is 5.

因此，最長的遞增子序列長度是5。

問題解決方法 (Problem Solution Approach)

Of course, in brute-force we can simply generate all increasing sequences and find the longest one. But it would take exponential time which is not a feasible solution. Hence, we choose Dynamic programming to solve.

當然，在蠻力作用下，我們可以簡單地生成所有遞增的序列并找到最長的序列。 但是，這將花費指數時間，這不是可行的解決方案。 因此，我們選擇動態規劃來解決。

We create a DP table to store longest increasing subsequence length.
It's intuitive that the minimum value would be 1 as each element represents the primitive sequence which is an increasing one.

我們創建一個DP表來存儲最長的遞增子序列長度。
直觀的是，最小值將為1，因為每個元素代表原始序列，該序列是遞增的。

So, the base value is 1.

因此，基準值為1。

Now,

現在，

Lis(i) = longest increasing subsequence starting from index 0 to index i

So,
To compute L(i) the recursion function is,

所以，
為了計算L(i) ，遞歸函數為

As, the base value is 1, for every index i, L(i) is at least 1.

這樣，對于每個索引i ，基值為1， L(i)至少為1。

1) Create the DP array, Lis[n]2) Initialize the DP array.for i=0 to n-1lis[i]=1;3) Now, to compute the Lis[i]for index i=1 to n-1 for previous index j=0 to i-1// if (arr[i],arr[j]) is inceasing sequenceif(lis[i]<lis[j]+1 && a[i]>a[j])lis[i]=lis[j]+1;end forend for

Initially DP table,

最初是DP表，

So, the maximum out of this is 5
Hence, LIS=5.

因此，最大數量為5
因此，LIS ＝ 5。

C++ implementation:

C ++實現：

#include <bits/stdc++.h> using namespace std;int max(int a, int b) {if (a > b)return a;elsereturn b; }int LIS(vector<int> a, int n) {int lis[n];//base casefor (int i = 0; i < n; i++)lis[i] = 1;//fill up tablefor (int i = 1; i < n; i++) {for (int j = 0; j < i; j++) {if (lis[i] < lis[j] + 1 && a[i] > a[j])lis[i] = lis[j] + 1;}}//return LISreturn *max_element(lis, lis + n); } int main() {int n, item;cout << "Sequence size:\n";scanf("%d", &n);//input the arrayvector<int> a;cout << "Input sequence:\n";for (int j = 0; j < n; j++) {scanf("%d", &item);a.push_back(item);}cout << "Length of longest incresing subsequence is: " << LIS(a, n) << endl;return 0; }

Output

輸出量

Sequence size: 10 Input sequence: 2 3 4 0 1 2 3 8 6 4 Length of longest incresing subsequence is: 5

翻譯自: https://www.includehelp.com/icp/longest-increasing-subsequence.aspx

最長遞增子序列 子串

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