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

---

• 经典DP。
• 第一种解法DP，定义f[ i ]为i位置时最长增长子序列长度，f[ i ] = max( f[ j ] + 1 ) if nums[ i ] > nums[ j ]。初始时f[ i ] = 1，目标函数为 max( f[ i ] )，注意不是f[ n ]。
• 第二种解法Patience sorting。假设把nums按照堆大小，分成不同堆。每个堆内都是一个增长子序列。定义tails为对应在各种堆大小时（i.e. 不同子序列长度），这些子序列中最小的尾元素。tails可证明是一个增长数组，因为若最小尾元素不增长的话，就不可能有一个更长的增长子序列。所以只要得出tails数组，它的长度就对应最长增长子序列长度。
  • 而要得到tails数组，就扫描一遍nums，然后判断需要添加到tails数组，还是更新它：
  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

class Solution:
    def lengthOfLIS1(self, nums: List[int]) -> int:
        if not nums:
            return 0
        
        dp = [1] * len(nums)
        
        for i in range(len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)
            
        # should be the maximum element of dp
        return max(dp)

    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums:
            return 0
        
        tails = []
        
        for num in nums:
            left, right = 0, len(tails) - 1
            
            # find the first index "left" where tails[left] >= num
            while left <= right:
                middle = (left + right) // 2
                if num <= tails[middle]:
                    right = middle - 1
                else:
                    left = middle + 1
            
            # if num is larger than all tails, append it
            if left == len(tails):
                tails.append(num)
            # if tails[left-1] < num <= tails[left], update tails[left]
            else:
                tails[left] = num
        
        return len(tails)