動機

讓我想起之前看有binary search的LIS，多看真的會有靈感

Problem

Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

 

Example 1:

Input: nums = [1,2,3,4,5]Output: trueExplanation: Any triplet where i < j < k is valid.

Example 2:

Input: nums = [5,4,3,2,1]Output: falseExplanation: No triplet exists.

Example 3:

Input: nums = [2,1,5,0,4,6]Output: trueExplanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

 

Constraints:

  • 1 <= nums.length <= 5 * 105
  • -231 <= nums[i] <= 231 - 1

 

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

Sol

因為只要找有沒有 A < B < C 所以 同時這個目標是遞迴的!!

所以就是從頭到尾一直更新剛好最接近的位置，同時判斷有沒有塞滿就好

class Solution:
    def increasingTriplet(self, nums: List[int], startFrom=0) -> bool:
        acc = [-math.inf]*3
        acc[0] = nums[0]
        for i in range(1,len(nums)):
            for j in range(3):
                if acc[j] == -math.inf or acc[j] >= nums[i]:
                    acc[j] = nums[i]
                    break
            if acc[2] != -math.inf:
                return True
        return False