📜  查找最长双调子序列的 C 程序

📅  最后修改于: 2022-05-13 01:54:30.104000             🧑  作者: Mango

查找最长双调子序列的 C 程序

给定一个包含 n 个正整数的数组 arr[0 … n-1],如果 arr[] 的子序列先递增,然后递减,则称为双调。编写一个函数数组为参数并返回最长双音子序列长度的函数。
一个按升序排序的序列被认为是双调的,降序部分为空。类似地,降序序列被认为是双调,增加的部分为空。
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来构造两个数组 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
", lbs( arr, n ) );
  return 0;
}


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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