最长的双子序列| DP-15 · GeeksForGeeks 中文系列教程

# 最长的双子序列| DP-15 > 原文： [https://www.geeksforgeeks.org/dynamic-programming-set-15-longest-bitonic-subsequence/](https://www.geeksforgeeks.org/dynamic-programming-set-15-longest-bitonic-subsequence/) 给定包含 n 个正整数的数组`arr[0…n-1]`，如果`arr[]`的[子序列](http://en.wikipedia.org/wiki/Subsequence)先增大然后减小，则称为 Bitonic。编写一个将数组作为参数并返回最长双子序列的长度的函数。按升序排序的序列被视为 Bitonic，而其递减部分为空。类似地，降序序列被视为 Bitonic，而升序部分为空。 **示例**： ``` 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[最长子序列（LIS）问题](https://www.geeksforgeeks.org/longest-increasing-subsequence-dp-3/)的变体。令输入数组为长度为`n`的`arr[]`。我们需要使用 [LIS 问题](https://www.geeksforgeeks.org/longest-increasing-subsequence-dp-3/)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++ ```cpp /* Dynamic Programming implementation of longest bitonic subsequence problem */ #include<stdio.h> #include<stdlib.h> /* 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; } ```