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.
Solution:
class Solution {
public int findLengthOfLCIS(int[] nums) {
// base case
if (nums.length <= 1){
return nums.length;
}
int cur = 1;
int result = 1;
for (int i = 1; i < nums.length; i++){
if (nums[i] > nums[i-1]){
cur = cur + 1;
result = Math.max(cur, result);
}else{
cur = 1;
}
}
return result;
}
}
/*
1 3 5 4 7
i
cur = 1
r = 3
if (nums[i] > nums[i-1]){
cur = cur+1;
r = Math.max(cur, r);
}else{
cur = 1;
}
return result;
*/
// TC: O(n)
// SC: O(1)