题目地址：https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/
题目描述：
Given an unsorted array of integers, find the number of longest increasing subsequence.
Example 1:
Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.
Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.
题目大意
这个题是最长递增子序列的变种。求最长的子序列有多少个。
解题方法
首先肯定还是使用dp去求。不过，我们得对dp的数组进行改进，我们在每个位置记录当前的LIS和能得到该LIS长度的子序列数目。在对每个位置进行计算的时候，我们都要找到该位置的LIS长度，并对能得到该长度的子序列的个数进行求和。
最后，我们需要对所有位置等于LIS长度的进行统计。
代码：
class Solution(object):
def findNumberOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dp, longest = [[1, 1] for _ in range(len(nums))], 1
for i in range(1, len(nums)):
curr_longest, count = 1, 0
for j in range(i):
if nums[j] < nums[i]:
curr_longest = max(curr_longest, dp[j][0] + 1)
for j in range(i):
if dp[j][0] == curr_longest - 1 and nums[j] < nums[i]:
count += dp[j][1]
dp[i] = [curr_longest, max(count, dp[i][1])]
longest = max(longest, curr_longest)
return sum([item[1] for item in dp if item[0] == longest])
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
DDKK.COM 弟弟快看-教程，程序员编程资料站，版权归原作者所有
本文经作者：负雪明烛 授权发布，任何组织或个人未经作者授权不得转发