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 {
int lengthOfLIS(List<int> nums) {
if (nums == null || nums.isEmpty) {
return 0;
}
List<int> dp = List.filled(nums.length + 1, 0);
// Prefill the dp table with a maximum value.
for (int i = 1; i < dp.length; i++) {
dp[i] = 1 << 31 - 1;
}
int left = 1;
int right = 1;
for (int curr in nums){
int start = left;
int end = right;
// Binary search, find the position to update in dp
while (start + 1 < end) {
int mid = start + (end - start) ~/ 2;
if (dp[mid] > curr) {
end = mid;
} else {
start = mid;
}
}
// Update the dp table
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] = curr;
right++;
}
}
return right;
}
}