题目：原题链接（中等）
标签：动态规划
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N 2 ) | O ( N ) | 1112ms (49.05%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def findNumberOfLIS(self, nums: List[int]) -> int:
size = len(nums)
if size <= 1:
return size
length = [0] * size
count = [1] * size
for j, num in enumerate(nums):
for i in range(j):
if nums[i] < nums[j]:
if length[i] >= length[j]:
length[j] = length[i] + 1
count[j] = count[i]
elif length[i] + 1 == length[j]:
count[j] += count[i]
longest = max(length)
return sum(n for i, n in enumerate(count) if length[i] == longest)
