题目

Given an integer array nums, return the number of longest increasing subsequences.

Notice that the sequence has to be strictly increasing.

Example 1:

Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: nums = [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.

Constraints:

• 1 <= nums.length <= 2000
• -106 <= nums[i] <= 106

思路

动态规划升级版。

代码

python版本：

class Solution:
    def findNumberOfLIS(self, nums: List[int]) -> int:
        dp = [[1, 1] for _ in range(len(nums))]
        longest = 1
        for i in range(len(nums)):
            maxn = 1
            count = 0
            for j in range(i):
                if nums[j] < nums[i]:
                    maxn = max(maxn, dp[j][0]+1)
            for j in range(i):
                if dp[j][0]+1 == maxn and nums[j] < nums[i]:
                    count += dp[j][1]
            dp[i] = [maxn, max(count, dp[i][1])]
            longest = max(longest, maxn)
        return sum([i[1] for i in dp if i[0] == longest])