问题描述：

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.

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算法：

从左到右扫描数组，放在新的数组中，并用Binary search 寻找当前元素要插入到新数组的位置，如果超出索引范围，数组扩张一位并放入当前元素，这样能保持新的数组的升序特性。否则，在插入位置上，将当前元素替换掉旧的元素，保存障碍信息。最终新的数组的长度就是结果。

import bisect
class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        seq = []
        for num in nums:
            insert_idx = bisect.bisect_left(seq, num) # duplicate 
            if insert_idx == len(seq):
                seq.append(num)
            else:
                seq[insert_idx] = num
        return len(seq)