Medium
Given an integer array nums
, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7]
is a subsequence of the array [0,3,1,6,2,2,7]
.
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500
-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n))
time complexity?
object Solution {
def lengthOfLIS(nums: Array[Int]): Int = {
if (nums == null || nums.isEmpty) {
return 0
}
val dp = Array.fill(nums.length + 1)(Int.MaxValue)
val left = 1
var right = 1
for (curr <- nums) {
var start = left
var end = right
while (start + 1 < end) {
val mid = start + (end - start) / 2
if (dp(mid) > curr) {
end = mid
} else {
start = mid
}
}
if (dp(start) > curr) {
dp(start) = curr
} else if (curr > dp(start) && curr < dp(end)) {
dp(end) = curr
} else if (curr > dp(end)) {
dp(end + 1) = curr
right += 1
}
}
right
}
}