# [LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列

2021年09月15日 阅读数：1

Given an unsorted array of integers, find the length of longest `continuous` increasing subsequence (subarray).html

Example 1:java

```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:python

```Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is , its length is 1. ```

Note: Length of the array will not exceed 10,000.数组

Java:url

```public int findLengthOfLCIS(int[] nums) {
int res = 0, cnt = 0;
for(int i = 0; i < nums.length; i++){
if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);
else cnt = 1;
}
return res;
}　　```

Python:code

```class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result, count = 0, 0
for i in xrange(len(nums)):
if i == 0 or nums[i-1] < nums[i]:
count += 1
result = max(result, count)
else:
count = 1
return result　　```

Python: wohtm

```class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0

n = len(nums)
dp =  * n
dp = 1
longest = 1
for i in xrange(1, n):
if nums[i] > nums[i-1]:
dp[i] = dp[i-1] + 1
else:
dp[i] = 1
longest = max(longest, dp[i])

return longest
```

C++:blog

```int findLengthOfLCIS(vector<int>& nums) {
int res = 0, cnt = 0;
for(int i = 0; i < nums.size(); i++){
if(i == 0 || nums[i-1] < nums[i]) res = max(res, ++cnt);
else cnt = 1;
}
return res;
}
```

[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列

[LeetCode] 673. Number of Longest Increasing Subsequence 最长递增序列的个数