代码随想录算法训练营第五十三天丨1143.最长公共子序列、1035. 不相交的线、53. 最大子数组和

1143. 最长公共子序列

LCS春节刷过了，算法导论书上的例题。

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        n1, n2 = len(text1), 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] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        
        return dp[n1][n2]

1035. 不相交的线

一模一样...如果不是这样的刷题顺序，这个题可能毫无头绪

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        n1, n2 = len(text1), 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] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        
        return dp[n1][n2]

53. 最大子数组和

这个题其实就是股票第一题..

也是算法导论分治法里面的例题

先给分治法解法：

分治：

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        # 辅助函数，用于找出跨越中点的最大子数组和
        def findMaxCrossingSubarray(nums, left, mid, right):
            # 左半边的最大子数组和
            left_sum = float('-inf')
            sum = 0
            for i in range(mid, left-1, -1):
                sum += nums[i]
                if sum > left_sum:
                    left_sum = sum
            
            # 右半边的最大子数组和
            right_sum = float('-inf')
            sum = 0
            for i in range(mid + 1, right + 1):
                sum += nums[i]
                if sum > right_sum:
                    right_sum = sum
            
            # 返回跨越中点的最大子数组和
            return left_sum + right_sum
        
        # 主递归函数
        def maxSubArrayRec(nums, left, right):
            # 基本情况
            if left == right:
                return nums[left]
            
            mid = (left + right) // 2
            # 分别找到左半边、右半边和跨越两边的最大子数组和
            left_sum = maxSubArrayRec(nums, left, mid)
            right_sum = maxSubArrayRec(nums, mid + 1, right)
            cross_sum = findMaxCrossingSubarray(nums, left, mid, right)
            
            # 返回这三个中的最大值
            return max(left_sum, right_sum, cross_sum)
        
        return maxSubArrayRec(nums, 0, len(nums) - 1)

动规：

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

今日总结：

今天的题其实挺难的，春节期间被折磨了两天。但是一旦不是一刷，就变简单了好多，可见刷题重要性。