LeetCode2021.3.25-132模式

O(n^3)的解法：（垃圾解法）

# class Solution(object):
#     def find132pattern(self, nums):
#         """
#         :type nums: List[int]
#         :rtype: bool
#         """
#         for forwardcount in range(len(nums)):
#             for lattercount in range(forwardcount+1, len(nums)):
#                 if nums[forwardcount] < nums[lattercount]:
#                     for midcount in range(lattercount+1, len(nums)):
#                         if nums[forwardcount] < nums[midcount] < nums[lattercount]:
#                             return True
#         return False

O（n^2）的解法：（via：负雪明烛：leetcode）

# class Solution(object):
#     def find132pattern(self, nums):
#         N = len(nums)
#         numsi = nums[0]
#         for j in range(1, N):
#             for k in range(N - 1, j, -1):
#                 # 从列表的两端进行维护
#                 if numsi < nums[k] < nums[j]:
#                     return True
#             numsi = min(numsi, nums[j])  # 始终维护numsi的小标小于nums[j]的下标
#         return False

单调栈的解法之后再进行补充