Python Leetcode之字符串相关问题

该类问题基本可以用动态规划进行求解，定义dp[i][j]/dp[i]表示以下表为 i -1 和j - 1 时的个数为dp[i][j]/dp[i].这种定义与之前的最长公共序列相关问题类似，i-1也表示的是以下标i-1结尾时的最长/最大公共序列为dp[i]。

判断子序列

讲解

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m = len(s)
        n = len(t)
        if m > n: return False
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s[i - 1] == t[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = dp[i][j - 1]
        return dp[-1][-1] == m

该题也可以使用双指针做法，当前两个字符相等时，则两个指针都进行加一，否则只对t的指针进行加一操作，代码如下：

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m, n = len(s), len(t)
        l, r = 0, 0
        while l < m and r < n:
            if s[l] == t[r]:
                l += 1
            r += 1
        return l == m 

不同的子序列

讲解

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        # dp[i][j]:S前i个，T前j个字符串组成的最多个数
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m + 1):
            dp[i][0] = 1
        for i in range(1, m + 1):
            for j in range(1, n + 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]

两个字符串的删除操作

讲解

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[float('inf')] * (n + 1) for _ in range(m + 1)]
        for i in range(n + 1):
            dp[0][i] = i
        for i in range(m + 1):
            dp[i][0] = i
        for i in range(1, m + 1):
            for j in range(1, n + 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] + 2, dp[i][j - 1] + 1, dp[i - 1][j] + 1)
        return dp[-1][-1]

编辑距离

讲解

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(n + 1):
            dp[0][i] = i
        for i in range(m + 1):
            dp[i][0] = i
        for i in range(1, m + 1):
            for j in range(1, n + 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] + 1)
        return dp[-1][-1]

另外一种解法：

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        import functools
        @functools.lru_cache(None)
        def helper(i, j):
            if i == len(word1) or j == len(word2):
                return len(word1) - i + len(word2) - j
            if word1[i] == word2[j]:
                return helper(i + 1, j + 1)
            else:
                inserted = helper(i, j + 1)
                deleted = helper(i + 1, j)
                replaced = helper(i + 1, j + 1)
                return min(inserted, deleted, replaced) + 1
        return helper(0, 0)