代码随想训练营第五十二天（Python）｜ 300.最长递增子序列、674. 最长连续递增序列、718. 最长重复子数组

300.最长递增子序列

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return len(nums)
        # dp 数组代表以 nums[i] 结尾的最长递增子序列长度为 dp[i]
        dp = [1] * len(nums)
        res = 1
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)
            res = max(dp[i], res)
        return res

674. 最长连续递增序列
1、贪心法

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return len(nums)
        max_res = 1
        cur = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i-1]:
                cur += 1
            else:
                cur = 1
            max_res = max(max_res, cur)
        return max_res

2、动态规划法

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return len(nums)
        # dp 数组代表 nums[i] 结尾的最长连续递增子序列为 dp[i]
        dp = [1] * len(nums)
        res = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i-1]:
                dp[i] = dp[i-1] + 1
            res = max(dp[i], res)
        return res

718. 最长重复子数组

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        # dp 数组代表以 i-1 结尾的 nums1 数组和 j-1 结尾的 nums2 数组的最长公共子数组长度为 dp[i][j]
        n = len(nums1)
        m = len(nums2)
        dp = [[0] * (m+1) for _ in range(n+1)]

        res = 0
        for i in range(1, n+1):
            for j in range(1, m+1):
                if nums1[i-1] == nums2[j-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                res = max(res, dp[i][j])
        return res