Leetcode 300. Longest Increasing Subsequence

https://leetcode.com/problems/longest-increasing-subsequence/

Medium

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: Output: 4 
Explanation: The longest increasing subsequence is , therefore the length is . [10,9,2,5,3,7,101,18][2,3,7,101]4

Note:

• 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?

---

• 經典DP。
• 第一種解法DP，定義f[ i ]爲i位置時最長增加子序列長度，f[ i ] = max( f[ j ] + 1 ) if nums[ i ] > nums[ j ]。初始時f[ i ] = 1，目標函數爲 max( f[ i ] )，注意不是f[ n ]。
• 第二種解法Patience sorting。假設把nums按照堆大小，分紅不一樣堆。每一個堆內都是一個增加子序列。定義tails爲對應在各類堆大小時（i.e. 不一樣子序列長度），這些子序列中最小的尾元素。tails可證實是一個增加數組，由於若最小尾元素不增加的話，就不可能有一個更長的增加子序列。因此只要得出tails數組，它的長度就對應最長增加子序列長度。
  • 而要獲得tails數組，就掃描一遍nums，而後判斷須要添加到tails數組，仍是更新它：
  1. if x is larger than all tails, append it, increase the size by 1
  2. if tails[i-1] < x <= tails[i], update tails[i]
• https://leetcode.com/problems/longest-increasing-subsequence/solution/
• https://leetcode.com/problems/longest-increasing-subsequence/discuss/74824/JavaPython-Binary-search-O(nlogn)-time-with-explanation
• https://en.wikipedia.org/wiki/Patience_sorting
• https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/LongestIncreasingSubsequence.pdf

 1 class Solution:
 2     def lengthOfLIS1(self, nums: List[int]) -> int:
 3         if not nums:
 4             return 0
 5         
 6         dp = [1] * len(nums)
 7         
 8         for i in range(len(nums)):
 9             for j in range(i):
10                 if nums[i] > nums[j]:
11                     dp[i] = max(dp[i], dp[j] + 1)
12             
13         # should be the maximum element of dp
14         return max(dp)
15 
16     def lengthOfLIS(self, nums: List[int]) -> int:
17         if not nums:
18             return 0
19         
20         tails = []
21         
22         for num in nums:
23             left, right = 0, len(tails) - 1
24             
25             # find the first index "left" where tails[left] >= num
26             while left <= right:
27                 middle = (left + right) // 2
28                 if num <= tails[middle]:
29                     right = middle - 1
30                 else:
31                     left = middle + 1
32             
33             # if num is larger than all tails, append it
34             if left == len(tails):
35                 tails.append(num)
36             # if tails[left-1] < num <= tails[left], update tails[left]
37             else:
38                 tails[left] = num
39         
40         return len(tails)

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