Solution
The question is pretty common and straightforward. We keep track of two variables, res and currLen. res stores the size of longest increasing subarray found yet and currLen stores the current length of increasing subsequence. Iterate through the array and if array[i] > array[i-1] then increase currLen else if array[i]<=array[i-1] then this element may be the starting point of another increasing subarray so currLen=1. At every step, res stores the maximum length of increasing subarray so res = max(res, currLen).
int findLengthOfLCIS(vector<int>& nums) {
// sanity check
if (nums.size()==0)
return 0;
int res=1, currLen=1;
for (int i=1; i<nums.size(); i++) {
if (nums[i]>nums[i-1])
currLen++;
else
currLen = 1;
res = max(res, currLen);
}
return res;
}
Time complexity: O(n)
Space complexity: O(1)