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?

最长连续递增子序列，非常典型的DP题目。目前有两种解法。普通DP是O(n^2)的复杂度，还有一个使用二分优化前一个的版本是O(nlogn)(见leetcode讨论）。

使用DP解，状态f[i]为以序列i为结尾的最长递增序列的长度，转换方程为f[i]=max(f[j]+1,nums[j-1] < nums[i-1])。注意这是一个非常经典的子状态定义方法，可以大大简化算法，如果f[i]定义为序列的前i 个字符中最长递增序列的长度，则f[j]到f[i]的转换将特别麻烦。这种的代码如下：

class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        res = [1]*len(nums)
        for i in xrange(1,len(nums)):
            for j in xrange(0,i):
                if nums[j] < nums[i]:
                    res[i] = max(res[i],res[j]+1)
        return max(res)