【代码随想录训练营】【Day 53】【动态规划-12】| Leetcode 115, 583, 72

【代码随想录训练营】【Day 53】【动态规划-12】| Leetcode 115, 583, 72

需强化知识点

题目

115. 不同的子序列

• 注意 python 数组反序用法 result [::-1]

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        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[m][n]
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583. 两个字符串的删除操作

• 注意 dp 数组初始化

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        # 0-i-1,0-j-1 的最小步数
        dp = [[0]*(n+1) for _ in range(m+1)]
        for i in range(1, m+1):
            dp[i][0] = i
        for j in range(1, n+1):
            dp[0][j] = j
        
        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(min(dp[i-1][j-1]+2, dp[i-1][j]+1), dp[i][j-1]+1)
        return dp[m][n]
        
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72. 编辑距离

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0] * (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]+1)
        
        return dp[len(word1)][len(word2)]
         
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