300. Longest Increasing Subsequence

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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

求数组的最长递增子序列。经典dp问题，类似Longest Common Subsequence。 这里其实也有O(nlogn)的方法 用binary search 。这个问题一开始可以被分解为recursive的子问题，一步一步优化就可以得到带有memorization的iterative解法。初始化dp[i] = 1，即一个元素的递增序列。 假设以i - 1结尾的subarray里的LIS为dp[i - 1]，那么我们要求以i结尾的subarray里的LIS，dp[i]的时候，要把这个新的元素和之前所有的元素进行比较，同时逐步比较dp[j] + 1与dp[i]，假如发现更长的序列，我们则更新dp[i] = dp[j] + 1，继续增加j进行比较。当i之前的元素全部便利完毕以后，我们得到了当前以i结尾的subarray里的LIS，就是dp[i]。

Time Complexity - O(n²)， Space Complexity - O(n²)。

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