673.Number of Longest Increasing Subsequence—最长递增子序列的个数

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.

思路：

举个栗子：nums=[1,4,3,5,9,8,10]

先实例化 dp 和tn为[1,1,1,1 ,1 ,1 ,1]，分别存放递增序列的最大长度 和

每个递增序列对应的子序列的个数。

通过循环比较大小使得dp里的数据更新为递增序列的最大长度[ 1, 2 ,2 ,3 ,4 ,4, 5 ]

同时使 tn 更新为[1，1， 1，2，2，2，4 ]

现在递增的最大长度是max_len是 5，

最后取出在 tn 中和最大长度对应的数，就是最长递增子序列的个数。

Python代码实现：

class Solution(object):
    def findNumberOfLIS(self, nums):
        n = len(nums)
        max_len = 1
        res = 0
        dp = [1] * n
        tn = [1] * n
        for i in range(1, n):
            for j in range(0, i):
                if nums[j] < nums[i] and dp[j] + 1 > dp[i]:
                    dp[i] = dp[j] + 1
                    tn[i] = tn[j]
                elif nums[j] < nums[i] and dp[j] + 1 == dp[i]:
                    tn[i] += tn[j]

            max_len = max(max_len, dp[i])
        for k in range(0, n):
            if dp[k] == max_len:
                res += tn[k]
        return res