Task02：动态规划

最新推荐文章于 2025-12-16 10:44:31 发布

原创最新推荐文章于 2025-12-16 10:44:31 发布 · 160 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#python

这篇博客探讨了动态规划在解决字符串问题中的应用，包括找到最长的回文子串、编辑距离算法以及一系列的打家劫舍问题，如打家劫舍和打家劫舍II。同时，还讨论了如何寻找最长连续递增序列和最长回文子序列的问题。

1.最长回文子串

class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        ans = ""
        # 枚举子串的长度 l+1
        for l in range(n):
            # 枚举子串的起始位置 i，这样可以通过 j=i+l 得到子串的结束位置
            for i in range(n):
                j = i + l
                if j >= len(s):
                    break
                if l == 0:
                    dp[i][j] = True
                elif l == 1:
                    dp[i][j] = (s[i] == s[j])
                else:
                    dp[i][j] = (dp[i + 1][j - 1] and s[i] == s[j])
                if dp[i][j] and l + 1 > len(ans):
                    ans = s[i:j+1]
        return ans

编辑距离

class Solution:
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        n = len(word1)
        m = len(word2)
        
        # 有一个字符串为空串
        if n * m == 0:
            return n + m
        
        # DP 数组
        D = [ [0] * (m + 1) for _ in range(n + 1)]
        
        # 边界状态初始化
        for i in range(n + 1):
            D[i][0] = i
        for j in range(m + 1):
            D[0][j] = j
        
        # 计算所有 DP 值
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                left = D[i - 1][j] + 1
                down = D[i][j - 1] + 1
                left_down = D[i - 1][j - 1] 
                if word1[i - 1] != word2[j - 1]:
                    left_down += 1
                D[i][j] = min(left, down, left_down)
        
        return D[n][m]

打家劫舍

class Solution:
    def rob(self, nums: List[int]) -> int:
        if not nums:
            return 0

        size = len(nums)
        if size == 1:
            return nums[0]
        
        dp = [0] * size
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        for i in range(2, size):
            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1])
        
        return dp[size - 1]

打家劫舍II

class Solution:
    def rob(self, nums: [int]) -> int:
        def my_rob(nums):
            cur, pre = 0, 0
            for num in nums:
                cur, pre = max(pre + num, cur), cur
            return cur
        return max(my_rob(nums[:-1]),my_rob(nums[1:])) if len(nums) != 1 else nums[0]

5.最长回文子序列

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s)
        maxL = -1
        dp = [[0]*n for _ in range(n)]

        for i in range(n):
            dp[i][i] = 1
        
        for i in range(n-1, -1, -1):
            for j in range(i+1, n):
                if s[i] == s[j]:
                    dp[i][j] = dp[i+1][j-1] + 2
                else:
                    dp[i][j] = max(dp[i][j-1], dp[i+1][j])
        
        return dp[0][n-1]

6.最长连续递增序列

class Solution(object):
    def findLengthOfLCIS(self, nums):
        ans = anchor = 0
        for i in range(len(nums)):
            if i and nums[i-1] >= nums[i]: anchor = i
            ans = max(ans, i - anchor + 1)
        return ans