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?

用DP的方法，每一个dp[i]代表从nums[0]到这个元素nums[i]的最长的inreasing subsequence的长度。O(N^2)的解法：

 1 class Solution(object):
 2     def lengthOfLIS(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         if not nums:
 8             return 0
 9         n = len(nums)
10         dp = [1] * n
11         
12         for i in range(0, n):
13             for j in range(i+1, n):
14                 if nums[j] > nums[i]:
15                     dp[j] = max(dp[i] + 1, dp[j])
16         
17         return max(dp)

300 Longest Increasing Subsequence