代码随想录第45天

最新推荐文章于 2025-12-20 09:26:15 发布

原创最新推荐文章于 2025-12-20 09:26:15 发布 · 951 阅读

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#python

115.不同的子序列

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        n1 = len(s)
        n2 = len(t)
        dp = [[0] * (n1 + 1) for _ in range(n2 + 1)]
        for j in range(n1 + 1):
            dp[0][j] = 1
        for i in range(1, n2 + 1):
            for j in range(1, n1 + 1):
                if t[i - 1] == s[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]  + dp[i][j - 1]
                else:
                    dp[i][j] = dp[i][j - 1]
        #print(dp)
        return dp[-1][-1]

583. 两个字符串的删除操作

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(m):
            for j in range(n):
                if word1[i] == word2[j]:
                    dp[i+1][j+1] = dp[i][j] + 1
                else:
                    dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
        return m + n - dp[m][n] * 2

72. 编辑距离

class Solution:
    def minDistance(self, word1: str, word2: str) -> 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]