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class Solution:def findLength(self, nums1: List[int], nums2: List[int]) -> int:# i-nums1的切片长度  j-nums2的切片长度# dp[i][j]表示 在nums1中以到i为止 在nums2中到j为止的相同的子数组长度 nums1[i]=nums2[j]# 遍历时记录结果值size1 = len(nums1)size2 = len(nums2)dp = [[0] * size2 for _ in range(size1)]res = 0for m in range(size1):if nums1[m] == nums2[0]:dp[m][0] = 1res = 1for n in range(size2):if nums2[n] == nums1[0]:dp[0][n] = 1res = 1for i in range(1, size1):for j in range(1, size2):if nums1[i] == nums2[j]:dp[i][j] = dp[i - 1][j - 1] + 1res = max(res, dp[i][j])return res

class Solution:def findLengthOfLCIS(self, nums: List[int]) -> int:#1.不定长滑动窗口 res=1temp=1for i in range(1,len(nums)):if nums[i]>nums[i-1]:temp+=1else:temp=1res=max(res,temp)return res

class Solution:def lengthOfLIS(self, nums: List[int]) -> int:# i-提供的数组长度size = len(nums)dp = [0] * sizedp[0] = 1for i in range(1, size):# 在i-1前面找dp值最大的一个temp = 0for j in range(i - 1, -1, -1):if nums[j] < nums[i]:temp = max(dp[j], temp)dp[i] = temp + 1return max(dp)