Question description
674. Longest continuous incremental sequence
Given an array of unsorted integers, find the longest andContinuously incremented subsequences, and return the length of the sequence.
Continuously incremented subsequencesCan be subscripted by twolandr（l < r）OK if for eachl <= i < r, allnums[i] < nums[i + 1], then subsequence[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]]It is a continuous incremental sub-sequence.
Example 1:
enter：nums = [1,3,5,4,7] Output：3 explain：The longest continuous incremental sequence is [1,3,5], The length is3。 although [1,3,5,7] It is also a subsequence of ascending order, But it's not continuous，because 5 and 7 In the original array 4 Separate。
Example 2:
enter：nums = [2,2,2,2,2] Output：1 explain：The longest continuous incremental sequence is [2], The length is1。
hint:
1 <= <= 10^4
-10^9 <= nums[i] <= 10^9
Idea Analysis
One traversal:
Maintain a value that records the current incremental length: temp = 1, because the minimum incremental growth is also 1
Update temp:
- When the current value is greater than the previous value, it is an incremental sequence, temp = temp + 1
- When the current value is less than or equal to the previous value, classify temp into the result list and reset temp = 1
When ended, the last temp is grouped into the result list
Returns the largest value in the result list.
Of course, you can also not use list storage. You just need to reset temp, compare the current temp and the largest temp_max, and then update temp_max. Use lists to help you understand more.
AC Code
class Solution:
def findLengthOfLCIS(self, nums: List[int]) -> int:
rev = []
ln = len(nums)
if ln == 0 or ln == 1:
return ln
temp = 1
for i in range(1, ln):
if nums[i]>nums[i-1]:
temp += 1
else:
(temp)
temp = 1
(temp)
return max(rev)
Don't use list:
class Solution:
def findLengthOfLCIS(self, nums: List[int]) -> int:
rev = 0
ln = len(nums)
if ln == 0 or ln == 1:
return ln
temp = 1
for i in range(1, ln):
if nums[i]>nums[i-1]:
temp += 1
else:
rev = temp if temp > rev else rev
temp = 1
rev = temp if temp > rev else rev
return rev
refer to
Drawing solution algorithm: 674. Longest continuous incremental sequence - Longest continuous incremental sequence - LeetCode ()
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