给定一个由 N 个整数组成的数组，找到给定序列的最长子序列的长度，使得该子序列的所有元素都严格按降序排序。
例子：
Input: arr[] = [15, 27, 14, 38, 63, 55, 46, 65, 85]
Output: 3
Explanation: The longest decreasing subsequence is {63, 55, 46}
Input: arr[] = {50, 3, 10, 7, 40, 80}
Output: 3
Explanation: The longest decreasing subsequence is {50, 10, 7}
这个问题可以用动态规划解决
最优子结构：
令 arr[0…n-1] 是输入数组，lds[i] 是 LDS 的长度，以索引 i 结束，这样 arr[i] 是 LDS 的最后一个元素。
然后， lds[i] 可以递归地写为：
lds[i] = 1 + max(lds[j]) where i > j > 0 and arr[j] > arr[i] or
lds[i] = 1, if no such j exists.
要找到给定数组的 LDS，我们需要返回max(lds[i])其中 n > i > 0。
C++
// CPP program to find the length of the
// longest decreasing subsequence
#include
using namespace std;
// Function that returns the length
// of the longest decreasing subsequence
int lds(int arr[], int n)
{
int lds[n];
int i, j, max = 0;
// Initialize LDS with 1 for all index
// The minimum LDS starting with any
// element is always 1
for (i = 0; i < n; i++)
lds[i] = 1;
// Compute LDS from every index
// in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] && lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
// Select the maximum
// of all the LDS values
for (i = 0; i < n; i++)
if (max < lds[i])
max = lds[i];
// returns the length of the LDS
return max;
}
// Driver Code
int main()
{
int arr[] = { 15, 27, 14, 38, 63, 55, 46, 65, 85 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Length of LDS is " << lds(arr, n);
return 0;
} Java
// Java program to find the
// length of the longest
// decreasing subsequence
import java.io.*;
class GFG
{
// Function that returns the
// length of the longest
// decreasing subsequence
static int lds(int arr[], int n)
{
int lds[] = new int[n];
int i, j, max = 0;
// Initialize LDS with 1
// for all index. The minimum
// LDS starting with any
// element is always 1
for (i = 0; i < n; i++)
lds[i] = 1;
// Compute LDS from every
// index in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] &&
lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
// Select the maximum
// of all the LDS values
for (i = 0; i < n; i++)
if (max < lds[i])
max = lds[i];
// returns the length
// of the LDS
return max;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 15, 27, 14, 38,
63, 55, 46, 65, 85 };
int n = arr.length;
System.out.print("Length of LDS is " +
lds(arr, n));
}
}
// This code is contributed by anuj_67.Python 3
# Python 3 program to find the length of
# the longest decreasing subsequence
# Function that returns the length
# of the longest decreasing subsequence
def lds(arr, n):
lds = [0] * n
max = 0
# Initialize LDS with 1 for all index
# The minimum LDS starting with any
# element is always 1
for i in range(n):
lds[i] = 1
# Compute LDS from every index
# in bottom up manner
for i in range(1, n):
for j in range(i):
if (arr[i] < arr[j] and
lds[i] < lds[j] + 1):
lds[i] = lds[j] + 1
# Select the maximum
# of all the LDS values
for i in range(n):
if (max < lds[i]):
max = lds[i]
# returns the length of the LDS
return max
# Driver Code
if __name__ == "__main__":
arr = [ 15, 27, 14, 38,
63, 55, 46, 65, 85 ]
n = len(arr)
print("Length of LDS is", lds(arr, n))
# This code is contributed by ita_cC#
// C# program to find the
// length of the longest
// decreasing subsequence
using System;
class GFG
{
// Function that returns the
// length of the longest
// decreasing subsequence
static int lds(int []arr, int n)
{
int []lds = new int[n];
int i, j, max = 0;
// Initialize LDS with 1
// for all index. The minimum
// LDS starting with any
// element is always 1
for (i = 0; i < n; i++)
lds[i] = 1;
// Compute LDS from every
// index in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] &&
lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
// Select the maximum
// of all the LDS values
for (i = 0; i < n; i++)
if (max < lds[i])
max = lds[i];
// returns the length
// of the LDS
return max;
}
// Driver Code
public static void Main ()
{
int []arr = { 15, 27, 14, 38,
63, 55, 46, 65, 85 };
int n = arr.Length;
Console.Write("Length of LDS is " +
lds(arr, n));
}
}
// This code is contributed by anuj_67.PHP
Javascript
输出 ：
Length of LDS is 3时间复杂度：O(n 2 )
辅助空间：O(n)
相关文章：https://www.geeksforgeeks.org/longest-increasing-subsequence/
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