[leetcode] 300. Longest Increasing Subsequence

题目

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]
Output: 1

Constraints:

• 1 <= nums.length <= 2500
• -104 <= nums[i] <= 104

Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

思路

1. 动态规划
2. 耐心排序

代码

python版本：

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = [0]*len(nums)
        for i in range(len(nums)):
            less = [dp[j] for j in range(i) if nums[j] < nums[i]]
            dp[i] = max(less)+1 if less else 1
        return max(dp)

# 耐心排序
class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        helper = []
        for n in nums:
            if not helper:
                helper.append([n])
            elif n<=helper[0][-1]:
                helper[0][-1]=n
            elif n>helper[-1][-1]:
                helper.append(helper[-1]+[n])
            else:
                l,r=0,len(helper)
                while l<r:
                    m=(l+r)//2
                    if helper[m][-1]<n:
                        l=m+1
                    else:
                        r=m
                helper[l][-1]=n
        return len(helper[-1])