Day 48 动态规划 part14

3道题目
1143. 最长公共子序列
1035. 不相交的线
53. 最大子数组和

解题理解

1143

设dp[i][j]为text10: i-1=text20: j-1的最长公共子序列。

class Solution:def longestCommonSubsequence(self, text1: str, text2: str) -> int:n1 = len(text1)n2 = len(text2)dp = [[0] * (n2 + 1) for _ in range(n1 + 1)]for i in range(1, n1 + 1):for j in range(1, n2 + 1):if text1[i - 1] == text2[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1else:dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])return dp[-1][-1]

1035

这道题几乎可以看作跟上一题一模一样，因为连线不交叉的前提条件是连线数字的相对顺序不变，所以还是一道求最长公共子序列的题。

class Solution:def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:n1 = len(nums1)n2 = len(nums2)dp = [[0] * (n2 + 1) for _ in range(n1 + 1)]res = 0for i in range(1, n1 + 1):for j in range(1, n2 + 1):if nums1[i - 1] == nums2[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1else:dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])if dp[i][j] > res:res = dp[i][j]return res

53

这道题之前用贪心做过，这次除了回顾了一下贪心算法，也用动规实现了下。设dp[i]为以i为结尾的最大子数组和，递推公式也比较好想，dp[i] = max(nums[i], dp[i - 1] + nums[i])

class Solution:def maxSubArray(self, nums: List[int]) -> int:if len(nums) == 1:return nums[0]n = len(nums)dp = [0] * ndp[0] = nums[0]res = dp[0]for i in range(1, n):dp[i] = max(nums[i], dp[i - 1] + nums[i])if res < dp[i]:res = dp[i]return res