The easiest method: $O(n^3)$
三层嵌套结构:

from typing import List


class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        if len(nums) < 3:
            return False
        i = 0
        while i < len(nums)-2:
            j=1 + 1
            while j < len(nums)-1:
                k= j + 1
                while k < len(nums):
                    if nums[i] < nums[j] and nums[j] < nums[k]:
                        return i,j,k
                    k += 1
                j+=1
            i+=1
        return False
                



solution = Solution()
solution.increasingTriplet([0,4,2,1,0,-1,-3])

Linear time complexity:
The most important point is that the second minimum number indicates there is a smaller number before it, we only need to find other number larger than this number.

A number larger than second number implies there are two smaller number in front.
A number larger than third number means there is three smaller number in front.

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        if len(nums)<3:
            return False
        min_number = float('inf')
        second_number = float('inf')
        for num in nums:
            if num <= min_number:
                min_number = num
            elif num <= second_number:
                second_number = num
            else: 
                return True 
        return False