[LeetCode]Longest Increasing Subsequence

原创于 2016-04-13 15:47:07 发布 · 365 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

38 篇文章

订阅专栏

本文介绍了一种解决最长递增子序列问题的方法，通过动态规划算法实现，复杂度为O(n²)，并提供了完整的Java代码实现。此外，文章还提出了一种更高效的O(n log n)的解决方案作为进阶挑战。

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
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?

题解：动态规划题目

code：

public class Solution {
    public int lengthOfLIS(int[] nums) {
        
        if(nums.length == 0)
            return 0;
        int max = 1;
        int[] dp = new int[nums.length];
        Arrays.fill(dp,1);
        for(int i=1; i<nums.length; i++)
        {
            for(int j=0; j<i; j++)
            {
                if(nums[i]> nums[j])
                {
                    if(dp[j]+1>dp[i])
                        dp[i] = dp[j]+1;
                }
            }
            if(max<dp[i])
                max = dp[i];
        }
        return max;
    }
}