LeetCode in Kotlin
674. Longest Continuous Increasing Subsequence
Easy
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
Solution
class Solution {
fun findLengthOfLCIS(nums: IntArray): Int {
var ans = 1
var count = 1
for (i in 0 until nums.size - 1) {
if (nums[i] < nums[i + 1]) {
count++
} else {
ans = max(count, ans)
count = 1
}
}
return max(ans, count)
}
private fun max(n1: Int, n2: Int): Int {
return Math.max(n1, n2)
}
}