【朱迪的LeetCode刷题笔记】】334. Increasing Triplet Subsequence #Medium #Python
本帖最后由 Judie 于 2023-6-4 11:24 编辑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 < nums. If no such indices exists, return false.
Example 1:
Input: nums =
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums =
Output: false
Explanation: No triplet exists.
Example 3:
Input: nums =
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums == 0 < nums == 4 < nums == 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?
Judy
Python taught by Mike
class Solution(object):
def increasingTriplet(self, nums):
"""
:type nums: List
:rtype: bool
"""
i = None
j = None
for n in nums:
if i == None or n < i:
i = n
elif j == None:
if n > i:
j = n
else:
continue
elif n > j:
return True
elif i < n < j:
j = n
return False
Sol1
Python
https://leetcode.com/problems/increasing-triplet-subsequence/solutions/78995/python-easy-o-n-solution/?envType=study-plan-v2&envId=leetcode-75
def increasingTriplet(nums):
first = second = float('inf')
for n in nums:
if n <= first:
first = n
elif n <= second:
second = n
else:
return True
return False
{:10_275:}
页:
[1]