题目描述
Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).
Example 1:
Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3.
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4.
Example 2:
Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1.
Note: Length of the array will not exceed 10,000.
题目大意
找出数组中最长连续递增子序列（子数组）
解题方法
动态规划
直接使用dp作为到某个位置的最长连续子序列。所以，如果当前的值比前一个值大，那么dp应该是前面的一个位置的数值+1，否则当前的值应该是1。另外需要注意的是当输入是空的时候，应该返回0.
class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums: return 0
N = len(nums)
dp = [1] * N
for i in range(1, N):
if nums[i] > nums[i - 1]:
dp[i] = dp[i - 1] + 1
return max(dp)
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空间压缩DP
在上面的做法中看出，每步的结果之和上面一步有关，所以可以优化空间复杂度。
class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
longest = 0
cur = 0
for i in range(len(nums)):
if i == 0 or nums[i] > nums[i - 1]:
cur += 1
longest = max(longest, cur)
else:
cur = 1
return longest
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 二刷的时候版本。
class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums: return 0
N = len(nums)
dp = 1
res = 1
for i in range(1, N):
if nums[i] > nums[i - 1]:
dp += 1
res = max(res, dp)
else:
dp = 1
return res
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