## 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 , its length is 1. ``````

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

## 解法一

### Python

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

### Java

``````class Solution {
public int findLengthOfLCIS(int[] nums) {
if (nums.length == 0) {
return 0;
}
int count = 1;
int result = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
count++;
} else {
result = Math.max(result, count);
count = 1;
}
}
result = Math.max(result, count);
return result;
}
}``````

### C++

``````class Solution {
public:
int findLengthOfLCIS(vector<int> &nums) {
if (nums.size() == 0) {
return 0;
}
int count = 1;
int result = 1;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] > nums[i - 1]) {
count++;
} else {
result = max(result, count);
count = 1;
}
}
result = max(result, count);
return result;
}
};``````