LeetCode-in-All
300. Longest Increasing Subsequence
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?
Solution
class Solution {
/**
* @param Integer[] $nums
* @return Integer
*/
public function lengthOfLIS($nums) {
if ($nums == null || count($nums) == 0) {
return 0;
}
$dp = array_fill(0, count($nums) + 1, PHP_INT_MAX);
// prefill the dp table
for ($i = 1; $i < count($dp); $i++) {
$dp[$i] = PHP_INT_MAX;
}
$left = 1;
$right = 1;
foreach ($nums as $curr) {
$start = $left;
$end = $right;
// binary search, find the one that is lower than curr
while ($start + 1 < $end) {
$mid = $start + (int)(($end - $start) / 2);
if ($dp[$mid] > $curr) {
$end = $mid;
} else {
$start = $mid;
}
}
// update our dp table
if ($dp[$start] > $curr) {
$dp[$start] = $curr;
} elseif ($curr > $dp[$start] && $curr < $dp[$end]) {
$dp[$end] = $curr;
} elseif ($curr > $dp[$end]) {
$dp[++$end] = $curr;
$right++;
}
}
return $right;
}
}