300. Longest Increasing Subsequence

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本文介绍了一种求解最长递增子序列(LIS)问题的动态规划算法，并提供了实现该算法的Java代码示例。该算法的时间复杂度为O(n²)，适用于未排序整数数组，通过动态规划计算出LIS的长度。

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问题链接：点击打开链接

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?

解题思路：定义一个dp数组，数组元素dp[i]表示以第i个元素结尾的递增序列的最大长度。DP算法，需要将前面元素的值都算出来

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