LeetCode-in-Rust
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
impl Solution {
pub fn length_of_lis(nums: Vec<i32>) -> i32 {
if nums.is_empty() {
return 0;
}
let mut dp = vec![i32::MAX; nums.len() + 1];
let (mut left, mut right) = (1, 1);
for &curr in nums.iter() {
let (mut start, mut end) = (left, right);
// Binary search to find the position to update
while start + 1 < end {
let mid = start + (end - start) / 2;
if dp[mid as usize] > curr {
end = mid;
} else {
start = mid;
}
}
// Update the dp array
if dp[start as usize] > curr {
dp[start as usize] = curr;
} else if curr > dp[start as usize] && curr < dp[end as usize] {
dp[end as usize] = curr;
} else if curr > dp[end as usize] {
dp[end as usize + 1] = curr;
right += 1;
}
}
right
}
}