给定一个包含 n 个正整数的数组 arr[0 … n-1]，如果 arr[] 的子序列首先增加，然后减少，则称为双音。编写一个函数，将数组作为参数并返回最长双音子序列的长度。
按递增顺序排序的序列被认为是双音，递减部分为空。类似地，降序序列被认为是双音的，递增部分为空。
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)来源：微软面试题
解决方案
此问题是标准最长递增子序列 (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 问题的动态规划解决方案构造两个数组 lis[] 和 lds[]。 lis[i] 存储以 arr[i] 结尾的最长递增子序列的长度。 lds[i] 存储了从 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 是从 0 到 n-1。
以下是上述动态规划解决方案的实现。
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)
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