leetcode-动态规划【子序列问题】

最新推荐文章于 2023-07-18 22:24:44 发布

cIlIegia_1234

最新推荐文章于 2023-07-18 22:24:44 发布

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分类专栏： leetcode 文章标签： leetcode 动态规划算法 python 数据结构

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本文深入探讨了动态规划在解决最长上升子序列、最长公共子序列、最长连续递增序列、最长重复子数组、最大子序和、编辑距离等经典问题中的应用。通过多个实例详细解析了不同类型的子序列问题以及编辑距离问题的解决方案，展示了动态规划在字符串处理和数组分析中的强大能力。

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目录

子序列（不连续）

300. 最长上升子序列

1143. 最长公共子序列

1035. 不相交的线

子序列（连续）

674. 最长连续递增序列

718. 最长重复子数组

53. 最大子序和

392. 判断子序列

115. 不同的子序列

583. 两个字符串的删除操作

72. 编辑距离

647. 回文子串

516. 最长回文子序列

子序列（不连续）

300. 最长递增子序列

class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # 以nums[i]结尾最长上升子序列的长度
        dp = [0 for _ in range(len(nums))]
        dp[0] = 1
        for i in range(1, len(nums)):
            max_len = 0
            for j in range(i):
                if nums[j] < nums[i]:
                    cur_len = dp[j] + 1
                else:
                    cur_len = 1
                max_len = max(max_len, cur_len)
            dp[i] = max_len
        return max(dp)

1143. 最长公共子序列

class Solution(object):
    def longestCommonSubsequence(self, text1, text2):
        """
        :type text1: str
        :type text2: str
        :rtype: int
        """
        # dp[i][j]:text1[0:i]与text2[0:j]最长子序列长度
        dp = [[0 for _ in range(len(text2))] for _ in range(len(text1))]
        for i in range(len(text1)):
            for j in range(len(text2)):
                if i == 0 and j == 0 and text1[0] == text2[0]:
                    dp[0][0] = 1
                if i == 0:
                    if dp[0][j-1] or text1[0] == text2[j]:
                        dp[0][j] = 1
                    continue
                if j == 0:
                    if dp[i-1][0] or text1[i] == text2[0]:
                        dp[i][0] = 1
                    continue
                if text1[i] == text2[j]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        return dp[-1][-1]

1035. 不相交的线

本题说是求绘制的最大连线数，其实就是求两个字符串的最长公共子序列的长度！和上一题的思路是一样的。

子序列（连续）

674. 最长连续递增序列

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # 以nums[i]为结尾的数组连续递增子序列长度为dp[i]
        dp = [0 for _ in range(len(nums))]
        dp[0] = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i-1]:
                dp[i] = dp[i-1] + 1
            else:
                dp[i] = 1
        return max(dp)

718. 最长重复子数组

class Solution(object):
    def findLength(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: int
        """
        # dp[i][j]：以nums1[i]和nums2[j]结尾的两个子序列最长重复子数组长度
        dp = [[0 for _ in range(len(nums2))] for _ in range(len(nums1))]
        ans = 0
        for i in range(len(nums1)):
            for j in range(len(nums2)):
                if nums1[i] != nums2[j]: dp[i][j] = 0
                else:
                    if i == 0 or j == 0: dp[i][j] = 1
                    else:
                        dp[i][j] = dp[i-1][j-1] + 1
                ans = max(ans, dp[i][j])
        return ans

53. 最大子序和

# 巧妙的解法
class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = -float('inf')
        ans = 0
        for i in range(len(nums)):
            ans += nums[i]
            if ans > result: result = ans
            if ans <= 0: ans = 0
        return result

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # dp[i]:以nums[i]为结尾的最大连续子数组和
        dp = [nums[0]]
        for i in range(1, len(nums)):
            dp.append(max(dp[i-1]+nums[i], nums[i]))
        return max(dp)

编辑距离

392. 判断子序列

# dp[i][j]:s[:j]是否为t[:i]的子序列 0/1
class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) == 0: return True
        if len(s) > len(t): return False
        # dp[i][j]:s[:j]是否为t[:i]的子序列 0/1
        dp = [[0 for _ in range(len(s))] for _ in range(len(t))]
        for i in range(len(t)):
            for j in range(len(s)):
                if i == 0 and j == 0:
                    if s[0] == t[0]: dp[i][j] = 1
                elif i == 0:
                    continue
                elif i < j:
                    continue
                else:
                    if t[i] == s[j]:
                        if j > 0: dp[i][j] = dp[i-1][j-1]
                        else: dp[i][j] = 1
                    else:
                        dp[i][j] = dp[i-1][j]
        return dp[-1][-1] == 1

# dp[i][j]:t[:i]和s[:j]的公共子序列长度
class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) == 0: return True
        if len(t) < len(s): return False
        # dp[i][j]:t[:i]和s[:j]的公共子序列长度
        dp = [[0 for _ in range(len(s))] for _ in range(len(t))]
        for i in range(len(t)):
            for j in range(len(s)):
                if t[i] == s[j]:
                    if i == 0 or j == 0: dp[i][j] = 1
                    else: dp[i][j] = dp[i-1][j-1] + 1
                else:
                    if i == 0 and j == 0: dp[i][j] = 0
                    elif i == 0: dp[i][j] = dp[i][j-1]
                    elif j == 0: dp[i][j] = dp[i-1][j]
                    else: dp[i][j] = max(dp[i][j-1], dp[i-1][j])
        return dp[-1][-1] == len(s)

115. 不同的子序列

class Solution(object):
    def numDistinct(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: int
        """
        # dp[i][j]:s[:i]中存在t[:j]的个数
        dp = [[0 for _ in range(len(t)+1)] for _ in range(len(s)+1)]
        # 当t为空序列时设置s包含一个t
        for i in range(len(s)+1):
            dp[i][0] = 1
        for i in range(1, len(s)+1):
            for j in range(1, len(t)+1):
                if s[i-1] == t[j-1]:
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
                else:
                    dp[i][j] = dp[i-1][j]
        return dp[-1][-1]

583. 两个字符串的删除操作

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        # dp[i][j]:在word1[:i]和word2[:j]中最长公共子序列长度
        dp = [[0 for _ in range(len(word2)+1)] for _ in range(len(word1)+1)]
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])
        max_len = dp[-1][-1]
        return len(word1) + len(word2) - 2*max_len

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        # dp[i][j]:word1[:i]和word2[:j]变成一样的序列所需要的操作数
        dp = [[0 for _ in range(len(word2)+1)] for _ in range(len(word1)+1)]
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j]+1, dp[i][j-1]+1, dp[i-1][j-1]+2)
        return dp[-1][-1]

72. 编辑距离

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        # dp[i][j]:word1[:i]和word2[:j]变成一样的序列所需要的操作数
        dp = [[0 for _ in range(len(word2)+1)] for _ in range(len(word1)+1)]
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    # 1. 插入：word1删除相当于word2插入，word2删除相当于word1插入
                    # 2. 删除：
                    # 2.1 word1删除：dp[i][j] = dp[i-1][j]+1
                    # 2.2 word2删除：dp[i][j] = dp[i][j-1]+1
                    # 3. 替换：dp[i][j] = dp[i-1][j-1]+1
                    dp[i][j] = min(dp[i-1][j]+1, dp[i][j-1]+1, dp[i-1][j-1]+1)
        return dp[-1][-1]

回文

647. 回文子串

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        count = 0
        # dp[i][j]:s[i:j]内的子串是否为回文子串
        dp = [[0 for _ in range(len(s))] for _ in range(len(s))]
        # 从左下角开始遍历
        for i in range(len(s)-1, -1, -1):
            for j in range(i, len(s)):
                if s[i] == s[j]:
                    if j-i <= 1:
                        dp[i][j] = 1
                        count += 1
                    else:
                        if dp[i+1][j-1]:
                            dp[i][j] = 1
                            count += 1
        return count

516. 最长回文子序列

class Solution(object):
    def longestPalindromeSubseq(self, s):
        """
        :type s: str
        :rtype: int
        """
        # dp[i][j]:s[i:j]内的最长回文子串长度
        dp = [[0 for _ in range(len(s))] for _ in range(len(s))]
        for i in range(len(s)):
            dp[i][i] = 1
        # 从左下角开始遍历
        for i in range(len(s)-1, -1, -1):
            for j in range(i, len(s)):
                if s[i] == s[j]:
                    if j-i <= 1: dp[i][j] = j-i+1
                    else: 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][-1]

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