# Longest Increasing Subsequence

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。

```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;
}```
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