[leetcode] 334. Increasing Triplet Subsequence @ python

最新推荐文章于 2022-07-22 03:27:20 发布

原创最新推荐文章于 2022-07-22 03:27:20 发布 · 242 阅读

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本文介绍了一种高效算法，用于检测一个未排序数组中是否存在长度为3的递增子序列。该算法通过定义两个变量first和second来追踪可能的递增序列，并在O(n)的时间复杂度和O(1)的空间复杂度下完成任务。

原题

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5]
Output: true
Example 2:

Input: [5,4,3,2,1]
Output: false

解法

定义两个变量: first, second表示递增序列的第一个和第二个变量. 遍历nums, 如果当前参数小于first, 更新first, 如果当前参数小于second, 更新second, 如果当前参数比first, second都大, 说明我们已找到一个匹配子序列.

Time: O(n)
Space: O(1)

代码

class Solution(object):
    def increasingTriplet(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        if len(nums) < 3:
            return False
        first = second = float('inf')
        for n in nums:
            if n <= first:
                first = n
            elif n <= second:
                second = n
            else:
                return True
        return False