最长递增子序列 (LIS) 问题是找到给定序列的最长子序列的长度，使子序列的所有元素按递增顺序排序。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}。
更多例子：
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}最优子结构：
令 arr[0..n-1] 是输入数组，L(i) 是 LIS 的长度，以索引 i 结束，这样 arr[i] 是 LIS 的最后一个元素。
然后，L(i) 可以递归地写为：
L(i) = 1 + max( L(j) ) 其中 0 < j < i 和 arr[j] < arr[i];要么
L(i) = 1，如果不存在这样的 j。
要找到给定数组的 LIS，我们需要返回 max(L(i)) 其中 0 < i < n。
因此，我们看到 LIS 问题满足最优子结构属性，因为主要问题可以使用子问题的解来解决。
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。
C++
/* A Naive C++ recursive implementation of LIS problem */
#include
using namespace std;
/* 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 */
int _lis(int arr[], int n, int* max_ref)
{
/* 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, max_ref);
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()
int lis(int arr[], int n)
{
// The max variable holds the result
int max = 1;
// The function _lis() stores its result in max
_lis(arr, n, &max);
// returns max
return max;
}
// Driver code
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Length of lis is " << lis(arr, n) << "\n";
return 0;
}
// This code is contributed by Shubhamsingh10 C
/* A Naive C recursive implementation of LIS problem */
#include
#include
/* 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 */
int _lis(int arr[], int n, int* max_ref)
{
/* 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, max_ref);
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()
int lis(int arr[], int n)
{
// The max variable holds the result
int max = 1;
// The function _lis() stores its result in max
_lis(arr, n, &max);
// returns max
return max;
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Length of lis is %d\n",
lis(arr, n));
return 0;
} 输出：
Length of lis is 5请参阅关于动态规划的完整文章 |设置 3(最长递增子序列)了解更多详情！
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