目录

学习目标

学习内容

 1143.最长公共子序列 

  1035.不相交的线  

  53. 最大子序和  动态规划 

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学习目标

学习内容

 1143.最长公共子序列 

1143. 最长公共子序列 - 力扣（LeetCode）https://leetcode.cn/problems/longest-common-subsequence/

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        n = len(text1)
        m = len(text2)
        dp = [[0]*(n+1)for _ in range(m+1)]
        for i in range(1,m+1):
            for j in range(1,n+1):
                if text1[j-1]==text2[i-1]:
                    dp[i][j] = dp[i-1][j-1]+1
                else:
                    dp[i][j]=max(dp[i][j-1],dp[i-1][j])
        #print(dp)
        return dp[-1][-1]

  1035.不相交的线  

1035. 不相交的线 - 力扣（LeetCode）https://leetcode.cn/problems/uncrossed-lines/

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        # 相当于最大公共子序列（不连续）
        # dp[i][j]表示text1[0:i-1]和text[0:j-1]的最长公共子序列
        # 递推公式 if text1[i-1]==text2[j-1]： dp[i][j] =  dp[i-1][j-1]+1 else  dp[i][j] = max(dp[i-1][j],dp[i][j-1])
        # 初始化 0
        # 递推公式 从左到右 从上到下

        n = len(nums1)
        m = len(nums2)
        dp = [[0]*(m+1)for _ in range(n+1)]
        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
                else:
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1])
        # print(dp)
        return dp[-1][-1]

  53. 最大子序和  动态规划 

53. 最大子数组和 - 力扣（LeetCode）https://leetcode.cn/problems/maximum-subarray/

import math
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        n = len(nums)
        if n==1:return nums[0]
        dp = [[-math.inf]*(n)for _ in range(2)]
        dp[1][0] = nums[0]
        res = -math.inf
        for i in range(1,n):
            dp[0][1] = max(dp[0][i-1],dp[1][i-1])
            dp[1][i] = max(0,dp[1][i-1])+nums[i]
            res = max(res,dp[0][i],dp[1][i])
        print(dp)
        return res

import math
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        n = len(nums)
        if n==1:return nums[0]
        a = -math.inf
        b = nums[0]
        res = -math.inf
        for i in range(1,n):
            a = max(a,b)
            b = max(0,b)+nums[i]
            res = max(res,a,b)
        return res