[leetcode] 334. Increasing Triplet Subsequence @ python

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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