【LeetCode】300. Longest Increasing Subsequence 解题报告（Python）

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本文链接： https://blog.csdn.net/ttinch/article/details/102673387

题目地址：https://leetcode.com/problems/longest-increasing-subsequence/

题目描述

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?

解法1：动态规划

$dp[i] = max(dp[j] + 1,1)$，其中 $j < i$且 $nums[j] < nums[i]$
算法复杂度为 $O(n^{2})$

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums:
            return 0
        dp = [1 for i in range(len(nums))]
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[j]+1, dp[i])
        return max(dp)

解法二：贪心算法+二分查找

具体思想可参考 https://blog.csdn.net/lw_power/article/details/80758674

复杂度 $O(nlogn)$：

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums:
            return 0
        tail = nums[0:1]
        for i in range(1, len(nums)):
            if nums[i] > tail[-1]:
                tail.append(nums[i])
            else:
                position = self.binarySearch(tail, nums[i])
                tail[position] = nums[i]
        return len(tail)

    # 查找nums中第一个比n大的元素
    def binarySearch(self, nums: List[int], n: int) -> int:
        # print(nums, n)
        left = 0
        right = len(nums) - 1
        while left < right:
            mid = (left+right)//2
            if nums[mid] == n:
                return mid
            elif nums[mid] > n:
                right = mid
            else:
                left = mid + 1
        if left == right:
            mid = left
        return mid