代码随想录算法训练营Day 52 | 动态规划part13 | 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

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300.最长递增子序列

题目链接

一、一维DP

class Solution(object):def lengthOfLIS(self, nums):""":type nums: List[int]:rtype: int"""if len(nums) <= 1:return len(nums)# dp[i]表示i之前包括i的以nums[i]结尾的最长递增子序列的长度dp=[1]*len(nums)for i in range(1,len(nums)):for j in range(i):if nums[j]<nums[i]:dp[i]=max(dp[i],dp[j]+1)return dp[-1]

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674. 最长连续递增序列

题目链接

一、贪心

class Solution(object):def findLengthOfLCIS(self, nums):""":type nums: List[int]:rtype: int"""maxcount =1count=1for i in range(len(nums)-1):  if nums[i+1]>nums[i]:count +=1      else: #不连续，count从头开始count =1maxcount = max(maxcount,count)return maxcount

二、动态规划

class Solution(object):def findLengthOfLCIS(self, nums):""":type nums: List[int]:rtype: int"""if len(nums) <= 1:return len(nums)# 动态规划dp=[1]*len(nums)res =1for i in range(1,len(nums)):if nums[i-1]<nums[i]:dp[i]=dp[i-1]+1res = max(res,dp[i])return res

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718. 最长重复子数组

题目链接

一、二维动态规划

class Solution(object):def findLength(self, nums1, nums2):""":type nums1: List[int]:type nums2: List[int]:rtype: int"""# dp[i][j] ：以下标i - 1为结尾的A，和以下标j - 1为结尾的B，最长重复子数组长度为dp[i][j]。 （特别注意： “以下标i - 1为结尾的A” 标明一定是 以A[i-1]为结尾的字符串 ）dp=[[0]*(len(nums2)+1) for _ in range(len(nums1)+1)]res=0# 全初始化为0# 递推公式for i in range(1,len(nums1)+1):for j in range(1,len(nums2)+1):if nums1[i-1]==nums2[j-1]:dp[i][j]=dp[i-1][j-1]+1res=max(res,dp[i][j])return res

二、转换为字符串判断

class Solution(object):def findLength(self, nums1, nums2):""":type nums1: List[int]:type nums2: List[int]:rtype: int"""s1 = "".join([chr(num) for num in nums1])s2 = "".join([chr(num) for num in nums2])n, res = len(s1), 0for i in range(n):for j in range(res+i+1, n+1):if s1[i:j] in s2:res = j - ielse:breakreturn res

三、一维DP

class Solution(object):def findLength(self, nums1, nums2):""":type nums1: List[int]:type nums2: List[int]:rtype: int"""# dp[i]：最长公共子数组长度dp=[0]*(len(nums2)+1)res=0for i in range(1,len(nums1)+1):for j in range(len(nums2), 0, -1):if nums1[i-1] == nums2[j-1]:dp[j] = dp[j-1] + 1else:dp[j]=0res=max(res,dp[j])return res