📜  最长递增子序列的Java程序

📅  最后修改于: 2021-09-22 09:57:02             🧑  作者: Mango

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}。
最长递增子序列

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。

/* 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,以下是大小为 4 的数组的递归树。 lis(n) 为我们提供了 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,有很多子问题被一次又一次地解决了。所以这个问题具有重叠子结构的性质,并且可以通过使用 Memoization 或 Tabulation 来避免相同子问题的重新计算。以下是 LIS 问题的列表实现。

/* 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!

如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程学生竞争性编程现场课程