最长公共序列、串问题总结

【最长公共子序列】

class Solution(object):
    def longestCommonSubsequence(self, text1, text2):
        """
        :type text1: str
        :type text2: str
        :rtype: int
        """
        m = len(text1)
        n = 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[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[m][n]

【最长公共子串】

class Solution:
    def LCS(self , str1: str, str2: str) -> str:
        #让str1为较长的字符串
        if len(str1) < len(str2): 
            str1, str2 = str2, str1
        res = ''
        max_len = 0
        #遍历str1的长度
        for i in range(len(str1)): 
            #查找是否存在
            if str1[i - max_len : i + 1] in str2: 
                res = str1[i - max_len : i + 1]
                max_len += 1
        return res

class Solution:
    def LCS(self , str1: str, str2: str) -> str:
        # write code here
        m = len(str1)
        n = len(str2)
        
        dp = [[0]*(n+1) for _ in range(m+1)]
        z = 0
        flag = 0
        
        for i in range(m):
            for j in range(n):
                if str1[i] == str2[j]:
                    dp[i+1][j+1] = dp[i][j] + 1
                    if dp[i+1][j+1] >z:
                        z = dp[i+1][j+1]
                        flag = i+1
        return str1[flag-z:flag]

【最长连续序列】

class Solution(object):
    def longestConsecutive(self, nums):
        hash_dict = dict()
        
        max_length = 0
        for num in nums:
            if num not in hash_dict:
                left = hash_dict.get(num - 1, 0)
                right = hash_dict.get(num + 1, 0)
                
                cur_length = 1 + left + right
                if cur_length > max_length:
                    max_length = cur_length
                
                hash_dict[num] = cur_length
                hash_dict[num - left] = cur_length
                hash_dict[num + right] = cur_length
                
        return max_length

【最长递增子序列】

class Solution:
    def LIS(self , arr: List[int]) -> int:
        # write code here
        
        dp = [1 for _ in range(len(arr))]
        res = 0
        for i in range(1,len(arr)):
            for j in range(i):
                if arr[i]>arr[j] and dp[i]<dp[j]+1:
                    dp[i] = dp[j] + 1
                    res = max(res,dp[i])
        return res

【最长回文子序列】

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

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

【最长回文子串】

class Solution:
    def getLongestPalindrome(self , A: str) -> int:
        # write code here
        n = len(A)
        if n < 2:
            return len(A)
        
        max_len = 1
        begin = 0
        # dp[i][j] 表示 s[i..j] 是否是回文串
        dp = [[False] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = True
        
        # 递推开始
        # 先枚举子串长度
        for L in range(2, n + 1):
            # 枚举左边界，左边界的上限设置可以宽松一些
            for i in range(n):
                # 由 L 和 i 可以确定右边界，即 j - i + 1 = L 得
                j = L + i - 1
                # 如果右边界越界，就可以退出当前循环
                if j >= n:
                    break
                    
                if A[i] != A[j]:
                    dp[i][j] = False 
                else:
                    if j - i < 3:
                        dp[i][j] = True
                    else:
                        dp[i][j] = dp[i + 1][j - 1]
                
                # 只要 dp[i][L] == true 成立，就表示子串 s[i..L] 是回文，此时记录回文长度和起始位置
                if dp[i][j] and j - i + 1 > max_len:
                    max_len = j - i + 1
                    begin = i
        return max_len

【最长无重复子串】

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        if len(s) <= 1: return len(s)

        char_map, left, ans = dict(), -1, 0
        for i, char in enumerate(s):
            if char not in char_map:
                char_map[char] = i
            else:
                left = max(left, char_map[char])
                char_map[char] = i
            ans = max(ans, i - left)
        
        return ans

【最长无重复子数组】

class Solution:
    def maxLength(self , arr: List[int]) -> int:
        # write code here
        
        dp = [0 for _ in range(len(arr)+1)]
        from collections import defaultdict
        temp = defaultdict(int)
        
        for i in range(1,len(arr)+1):

            dp[i] = min(i-temp[arr[i-1]],dp[i-1]+1)
            temp[arr[i-1]] = i
        return max(dp)