**Longest Increasing Subsequence**

Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

**Example**

For [5, 4, 1, 2, 3], the LIS is [1, 2, 3], return 3

For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4

**Challenge**

Time complexity O(n^2) or O(nlogn)

**Clarification**

What’s the definition of longest increasing subsequence?

* The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence’s elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.

* https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Solution: DP O(n^2)

lis[i]表示结尾为nums[i]的子序列的最长长度。

所以lis[i] =for all the j<i && nums[j]<=nums[i] max(lis[j])+1

其实还有可以优化为lis[i] = lis[j] +1 where nums[j] 是比nums[i]小的数里最大的一个

public int longestIncreasingSubsequence(int[] nums) { if (nums == null || nums.length == 0) { return 0; } //lis[i] means the lis of nums[0-i] which i is the last element selected in the lis int[] lis = new int[nums.length]; int maxLis = 0; for (int i = 0; i < nums.length; i++) { lis[i] = 1; for (int j = 0; j < i; j++) { if (nums[i] >= nums[j]) { lis[i] = Math.max(lis[j] + 1, lis[i]); } } maxLis = Math.max(maxLis, lis[i]); } return maxLis; }