334. Increasing Triplet Subsequence - python3
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334. Increasing Triplet Subsequence
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.class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
for i in range(len(nums)-2):
if nums[i] < nums[i+1] and nums[i+1] < nums[i+2]:
return True
return False
最初は3回連続でいいと思ってたけど、違う
My Answer 1: Accepted (Runtime: 3968 ms - 7.99% / Memory Usage: 14.8MB - 49.89%)
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
for i in range(len(nums)-2):
numlist = []
for j in range(i+1, len(nums)):
if nums[i] < nums[j]:
numlist.append(nums[j])
for j in range(len(numlist)-1):
if numlist[j] < numlist[j+1]:
return True
return False
7運行時の主人公の中にFor Moon.^^
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
for i in range(len(nums)-2):
if nums[i] < nums[i+1] and nums[i+1] < nums[i+2]:
return True
return Falseclass Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
for i in range(len(nums)-2):
numlist = []
for j in range(i+1, len(nums)):
if nums[i] < nums[j]:
numlist.append(nums[j])
for j in range(len(numlist)-1):
if numlist[j] < numlist[j+1]:
return True
return Falsenumlist[j] < numlist[j+1]がある場合、True Return Time Complexity:O(n) and Space:O(1)
Solution 1: Runtime: 56 ms - 57.56% / Memory Usage: 14.7 MB - 93.66%
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
f_min = float('inf')
s_min = float('inf')
for i in range(len(nums)):
if nums[i] <= f_min:
f_min = nums[i]
elif nums[i] <= s_min:
s_min = nums[i]
else:
return True
return FalseReference
この問題について(334. Increasing Triplet Subsequence - python3), 我々は、より多くの情報をここで見つけました https://velog.io/@jsh5408/334.-Increasing-Triplet-Subsequence-python3テキストは自由に共有またはコピーできます。ただし、このドキュメントのURLは参考URLとして残しておいてください。
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