最长增长子序列(LIS)问题是找到给定序列的最长子序列的长度，以使子序列的所有元素都按升序排序。例如，{10、22、9、33、21、50、41、60、80}的LIS长度为6，LIS为{10、22、33、50、60、80}。 
更多示例：
Input : arr[] = {3, 10, 2, 1, 20}
Output : Length of LIS = 3
The longest increasing subsequence is 3, 10, 20
Input : arr[] = {3, 2}
Output : Length of LIS = 1
The longest increasing subsequences are {3} and {2}
Input : arr[] = {50, 3, 10, 7, 40, 80}
Output : Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}
最佳子结构：
令arr [0..n-1]为输入数组，令L(i)为以索引i结尾的LIS的长度，以使arr [i]为LIS的最后一个元素。
然后，L(i)可以递归写为：
L(i)= 1 + max(L(j))其中0
为了找到给定数组的LIS，我们需要返回max(L(i))，其中0 因此，我们看到LIS问题满足最佳子结构属性，因为可以使用子问题的解决方案来解决主要问题。
以下是LIS问题的简单递归实现。它遵循上面讨论的递归结构。
/* A Naive Java Program for LIS Implementation */
class LIS {
static int max_ref; // stores the LIS
/* To make use of recursive calls, this function must return
two things:
1) Length of LIS ending with element arr[n-1]. We use
max_ending_here for this purpose
2) Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result */
static int _lis(int arr[], int n)
{
// base case
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS ending with arr[n-1]
int res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0], arr[1] ...
arr[n-2]. If arr[i-1] is smaller than arr[n-1], and
max ending with arr[n-1] needs to be updated, then
update it */
for (int i = 1; i < n; i++) {
res = _lis(arr, i);
if (arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall max. And
// update the overall max if needed
if (max_ref < max_ending_here)
max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
static int lis(int arr[], int n)
{
// The max variable holds the result
max_ref = 1;
// The function _lis() stores its result in max
_lis(arr, n);
// returns max
return max_ref;
}
// driver program to test above functions
public static void main(String args[])
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.length;
System.out.println("Length of lis is "
+ lis(arr, n) + "\n");
}
}
/*This code is contributed by Rajat Mishra*/
输出：
Length of lis is 5
重叠子问题：
考虑到上述实现，下面是大小为4的数组的递归树。lis(n)给出了arr []的LIS长度。
lis(4)
/ |
lis(3) lis(2) lis(1)
/ /
lis(2) lis(1) lis(1)
/
lis(1)
我们可以看到，有许多子问题一次又一次地得到解决。因此，此问题具有“重叠子结构”属性，可以通过使用“记忆化”或“制表”来避免相同子问题的重新计算。以下是LIS问题的列表化实施。
/* Dynamic Programming Java implementation of LIS problem */
class LIS {
/* lis() returns the length of the longest increasing
subsequence in arr[] of size n */
static int lis(int arr[], int n)
{
int lis[] = new int[n];
int i, j, max = 0;
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute optimized LIS values in bottom up manner */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
public static void main(String args[])
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.length;
System.out.println("Length of lis is " + lis(arr, n) + "\n");
}
}
/*This code is contributed by Rajat Mishra*/
输出：
Length of lis is 5
请参考有关动态编程的完整文章。设置3(最长递增子序列)以获取更多详细信息！