Leetcode 300 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?

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：

 class Solution(object):
     def lengthOfLIS(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         if not nums:
             return 0
         n = len(nums)
         dp = [1] * n

         for i in range(0, n):
             for j in range(i+1, n):
                 if nums[j] > nums[i]:
                     dp[j] = max(dp[i] + 1, dp[j])

         return max(dp)

解法二 O(N logN)

本题可以用一个dp list来维护LIS的solution。对于nums里面的数跑一个for loop。如果遇到dp = []，需要把当前的num填到dp list里面。如果发现num大于dp list的最后（也就是最大的那个数）就append上去。如果num在[dp[0], dp[-1]]区间内，用binary search找到num应该在的位置去替换相应的数。这么做的目的是，例如上图中用5替换了7，假如5后面有6,7两个数的话，就会取得一个更长的subsequence。注意mid和num比较时使用的是>还是>=。这回区间头是移动到mid+1还是mid。

 def lengthOfLIS(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         if not nums:
             return 0
         size = len(nums)
         dp = []
         for x in range(size):
             if not dp or dp[-1] < nums[x]:
                 dp.append(nums[x])
             low, high = 0, len(dp) - 1
             while low < high:
                 mid = (low + high)/2
                 if dp[mid] < nums[x]:
                     low = mid + 1
                 else:
                     high = mid
             dp[high] = nums[x]

         return len(dp)