📜  最长的双子序列| DP-15

📅  最后修改于: 2021-05-08 16:50:14             🧑  作者: Mango

给定包含n个正整数的数组arr [0…n-1],如果arr []的子序列先增大然后减小,则称为Bitonic。写一个函数,采用一个数组作为参数,并返回最长双调子序列的长度。
按升序排序的序列被视为Bitonic,而降序部分为空。类似地,降序序列被认为是Bitonic,升序部分为空。
例子:

Input arr[] = {1, 11, 2, 10, 4, 5, 2, 1};
Output: 6 (A Longest Bitonic Subsequence of length 6 is 1, 2, 10, 4, 2, 1)

Input arr[] = {12, 11, 40, 5, 3, 1}
Output: 5 (A Longest Bitonic Subsequence of length 5 is 12, 11, 5, 3, 1)

Input arr[] = {80, 60, 30, 40, 20, 10}
Output: 5 (A Longest Bitonic Subsequence of length 5 is 80, 60, 30, 20, 10)

资料来源:Microsoft面试问题

解决方案
此问题是标准最长增加子序列(LIS)问题的变体。令输入数组为长度为n的arr []。我们需要使用LIS问题的动态编程解决方案构造两个数组lis []和lds []。 lis [i]存储以arr [i]结尾的最长增加子序列的长度。 lds [i]存储从arr [i]开始的最长递减子序列的长度。最后,我们需要返回lis [i] + lds [i] – 1的最大值,其中i从0到n-1。
以下是上述动态编程解决方案的实现。

C++
/* Dynamic Programming implementation of longest bitonic subsequence problem */
#include
#include
 
/* lbs() returns the length of the Longest Bitonic Subsequence in
    arr[] of size n. The function mainly creates two temporary arrays
    lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
 
    lis[i] ==> Longest Increasing subsequence ending with arr[i]
    lds[i] ==> Longest decreasing subsequence starting with arr[i]
*/
int lbs( int arr[], int n )
{
   int i, j;
 
   /* Allocate memory for LIS[] and initialize LIS values as 1 for
      all indexes */
   int *lis = new int[n];
   for (i = 0; i < n; i++)
      lis[i] = 1;
 
   /* Compute LIS values from left to right */
   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;
 
   /* Allocate memory for lds and initialize LDS values for
      all indexes */
   int *lds = new int [n];
   for (i = 0; i < n; i++)
      lds[i] = 1;
 
   /* Compute LDS values from right to left */
   for (i = n-2; i >= 0; i--)
      for (j = n-1; j > i; j--)
         if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
            lds[i] = lds[j] + 1;
 
 
   /* Return the maximum value of lis[i] + lds[i] - 1*/
   int max = lis[0] + lds[0] - 1;
   for (i = 1; i < n; i++)
     if (lis[i] + lds[i] - 1 > max)
         max = lis[i] + lds[i] - 1;
   return max;
}
 
/* Driver program to test above function */
int main()
{
  int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5,
              13, 3, 11, 7, 15};
  int n = sizeof(arr)/sizeof(arr[0]);
  printf("Length of LBS is %d\n", lbs( arr, n ) );
  return 0;
}


Java
/* Dynamic Programming implementation in Java for longest bitonic
   subsequence problem */
import java.util.*;
import java.lang.*;
import java.io.*;
 
class LBS
{
    /* lbs() returns the length of the Longest Bitonic Subsequence in
    arr[] of size n. The function mainly creates two temporary arrays
    lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
 
    lis[i] ==> Longest Increasing subsequence ending with arr[i]
    lds[i] ==> Longest decreasing subsequence starting with arr[i]
    */
    static int lbs( int arr[], int n )
    {
        int i, j;
 
        /* Allocate memory for LIS[] and initialize LIS values as 1 for
            all indexes */
        int[] lis = new int[n];
        for (i = 0; i < n; i++)
            lis[i] = 1;
 
        /* Compute LIS values from left to right */
        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;
 
        /* Allocate memory for lds and initialize LDS values for
            all indexes */
        int[] lds = new int [n];
        for (i = 0; i < n; i++)
            lds[i] = 1;
 
        /* Compute LDS values from right to left */
        for (i = n-2; i >= 0; i--)
            for (j = n-1; j > i; j--)
                if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
                    lds[i] = lds[j] + 1;
 
 
        /* Return the maximum value of lis[i] + lds[i] - 1*/
        int max = lis[0] + lds[0] - 1;
        for (i = 1; i < n; i++)
            if (lis[i] + lds[i] - 1 > max)
                max = lis[i] + lds[i] - 1;
 
        return max;
    }
 
    public static void main (String[] args)
    {
        int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5,
                    13, 3, 11, 7, 15};
        int n = arr.length;
        System.out.println("Length of LBS is "+ lbs( arr, n ));
    }
}


Python
# Dynamic Programming implementation of longest bitonic subsequence problem
"""
lbs() returns the length of the Longest Bitonic Subsequence in
arr[] of size n. The function mainly creates two temporary arrays
lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
 
lis[i] ==> Longest Increasing subsequence ending with arr[i]
lds[i] ==> Longest decreasing subsequence starting with arr[i]
"""
 
def lbs(arr):
    n = len(arr)
 
 
    # allocate memory for LIS[] and initialize LIS values as 1
    # for all indexes
    lis = [1 for i in range(n+1)]
 
    # Compute LIS values from left to right
    for i in range(1 , n):
        for j in range(0 , i):
            if ((arr[i] > arr[j]) and (lis[i] < lis[j] +1)):
                lis[i] = lis[j] + 1
 
    # allocate memory for LDS and initialize LDS values for
    # all indexes
    lds = [1 for i in range(n+1)]
     
    # Compute LDS values from right to left
    for i in reversed(range(n-1)): #loop from n-2 downto 0
        for j in reversed(range(i-1 ,n)): #loop from n-1 downto i-1
            if(arr[i] > arr[j] and lds[i] < lds[j] + 1):
                lds[i] = lds[j] + 1
 
 
    # Return the maximum value of (lis[i] + lds[i] - 1)
    maximum = lis[0] + lds[0] - 1
    for i in range(1 , n):
        maximum = max((lis[i] + lds[i]-1), maximum)
     
    return maximum
 
# Driver program to test the above function
arr =  [0 , 8 , 4, 12, 2, 10 , 6 , 14 , 1 , 9 , 5 , 13,
        3, 11 , 7 , 15]
print "Length of LBS is",lbs(arr)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)


C#
/* Dynamic Programming implementation in
   C# for longest bitonic subsequence problem */
using System;
 
class LBS {
     
    /* lbs() returns the length of the Longest Bitonic Subsequence in
    arr[] of size n. The function mainly creates two temporary arrays
    lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
 
    lis[i] ==> Longest Increasing subsequence ending with arr[i]
    lds[i] ==> Longest decreasing subsequence starting with arr[i]
    */
    static int lbs(int[] arr, int n)
    {
        int i, j;
 
        /* Allocate memory for LIS[] and initialize
           LIS values as 1 for all indexes */
        int[] lis = new int[n];
        for (i = 0; i < n; i++)
            lis[i] = 1;
 
        /* Compute LIS values from left to right */
        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;
 
        /* Allocate memory for lds and initialize LDS values for
            all indexes */
        int[] lds = new int[n];
        for (i = 0; i < n; i++)
            lds[i] = 1;
 
        /* Compute LDS values from right to left */
        for (i = n - 2; i >= 0; i--)
            for (j = n - 1; j > i; j--)
                if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
                    lds[i] = lds[j] + 1;
 
        /* Return the maximum value of lis[i] + lds[i] - 1*/
        int max = lis[0] + lds[0] - 1;
        for (i = 1; i < n; i++)
            if (lis[i] + lds[i] - 1 > max)
                max = lis[i] + lds[i] - 1;
 
        return max;
    }
     
    // Driver code
    public static void Main()
    {
        int[] arr = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5,
                                      13, 3, 11, 7, 15 };
        int n = arr.Length;
        Console.WriteLine("Length of LBS is " + lbs(arr, n));
    }
}
 
// This code is contributed by vt_m.


PHP
 Longest Increasing subsequence
              ending with arr[i]
   lds[i] ==> Longest decreasing subsequence
              starting with arr[i]
*/
function lbs(&$arr, $n)
{
 
    /* Allocate memory for LIS[] and initialize
       LIS values as 1 for all indexes */
    $lis = array_fill(0, $n, NULL);
    for ($i = 0; $i < $n; $i++)
        $lis[$i] = 1;
     
    /* Compute LIS values from left to right */
    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;
     
    /* Allocate memory for lds and initialize
       LDS values for all indexes */
    $lds = array_fill(0, $n, NULL);
    for ($i = 0; $i < $n; $i++)
        $lds[$i] = 1;
     
    /* Compute LDS values from right to left */
    for ($i = $n - 2; $i >= 0; $i--)
        for ($j = $n - 1; $j > $i; $j--)
            if ($arr[$i] > $arr[$j] &&
                $lds[$i] < $lds[$j] + 1)
                $lds[$i] = $lds[$j] + 1;
     
    /* Return the maximum value of
       lis[i] + lds[i] - 1*/
    $max = $lis[0] + $lds[0] - 1;
    for ($i = 1; $i < $n; $i++)
        if ($lis[$i] + $lds[$i] - 1 > $max)
            $max = $lis[$i] + $lds[$i] - 1;
    return $max;
}
 
// Driver Code
$arr = array(0, 8, 4, 12, 2, 10, 6, 14,
             1, 9, 5, 13, 3, 11, 7, 15);
$n = sizeof($arr);
echo "Length of LBS is " . lbs( $arr, $n );
 
// This code is contributed by ita_c
?>


Javascript


输出:

Length of LBS is 7

时间复杂度:O(n ^ 2)
辅助空间:O(n)