Problem
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.
Solution
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int len = nums.size();
int longest = INT_MIN;
if(len==0)return 0;
int pre=nums[0];
int count=1;
for(int i=1;i<len;i++){
if(nums[i]>pre){
count++;
pre=nums[i];
}else{
longest=max(longest,count);
pre=nums[i];
count=1;
}
}
longest=max(longest,count);
return longest;
}
};
static auto speedup = [](){
ios::sync_with_stdio(false);
cin.tie(nullptr);
return nullptr;
}();
image-20190228092013276-1316813.png
