334. Increasing Triplet Subsequence【M】【69】

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.

Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:

Given

[1, 2, 3, 4, 5]

,

return

true

.

Given

[5, 4, 3, 2, 1]

,

return

false

.

Credits:

Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

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就是维护一个大小为3的窗口

low代表最小的，mid代表中间大小的

如果当前数，大于mid，那肯定满足条件

这里面有个trick，就是low的位置不一定在mid前面，它可以代表有可能成为最小的的候选。

class Solution(object):
def increasingTriplet(self, nums):
if len(nums)<3:
return False

mid=sys.maxint
low=nums[0]
for i in nums[1:]:
#print low,mid
if i > mid:
return True
if i > low:
mid = min(mid,i)
low = min(low,i)
return False