674. Longest Continuous Increasing Subsequence
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 <= 104
- -109 <= nums[i] <= 109
class Solution {
public int findLengthOfLCIS(int[] nums) {
if(nums == null || nums.length == 0){
return 0;
}
int n = nums.length;
int[] dp = new int[n];
int max = 1;
dp[0] = 1;
for(int i=1;i<n;i++){
if(nums[i] > nums[i-1]){
dp[i] = dp[i-1] + 1;
}else{
dp[i] = 1;
}
max = Math.max(max, dp[i]);
}
return max;
}
}