Problem
Given an unsorted array of integers nums, return **the length of the longest *continuous increasing subsequence* (i.e. subarray)**. The subsequence must be *strictly* increasing.
A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].
Example 1:
Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.
Example 2:
Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.
Constraints:
1 <= nums.length <= 10^4-10^9 <= nums[i] <= 10^9
Solution (Java)
class Solution {
public int findLengthOfLCIS(int[] nums) {
int ans = 1;
int count = 1;
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] < nums[i + 1]) {
count++;
} else {
ans = max(count, ans);
count = 1;
}
}
return max(ans, count);
}
private int max(int n1, int n2) {
return Math.max(n1, n2);
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).