Given an unsorted array of integers, find the length of longest increasing subsequence.
For example,
Given
The longest increasing subsequence is
Given
[10, 9, 2, 5, 3, 7, 101, 18],The longest increasing subsequence is
[2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?
Understand the problem:A sequence DP problem.
-- dp[n + 1], where dp[i] means the length of the LIS, including the ith character.
-- Initial state: dp[0] = 0, dp[i] = 1, i >= 1
-- Transit function: for j from [0, i), if (nums[i - 1] > nums[j - 1]), dp[i] = Math.max(dp[i], dp[j] + 1);
-- Final state: the max in dp[i].
Code (Java):
public class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int[] dp = new int[nums.length + 1];
        for (int i = 1; i < nums.length + 1; i++) {
            dp[i] = 1;
        }
        
        int maxLen = 1;
        
        for (int i = 1; i <= nums.length; i++) {
            for (int j = 1; j < i; j++) {
                if (nums[i - 1] > nums[j - 1]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            
            maxLen = Math.max(maxLen, dp[i]);
        }
        
        return maxLen;
    }
}
Follow-up: Could you solve it in O(nlogn) time?
https://leetcode.com/discuss/67609/short-java-solution-using-dp-o-n-log-n
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