【代码随想录算法训练营】第56天 | 第九章 动态规划（十六）+ 复习第28天 第六章 回溯（四）

主要内容

编辑距离问题

动态规划题目

583. 两个字符串的删除操作

思路分析

题目可转变为求最长公共子序列长度

代码

class Solution:
    # 求最长公共子序列长度
    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(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] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        return m + n - 2*dp[-1][-1]
1
2
3
4
5
6
7
8
9
10
11
12

72. 编辑距离

思路分析

1、含义：dp[i][j]表示以word1[i-1] word2[j-1]结尾的单词的最近编辑距离
2、递推：
① word1[i-1] == word2[j-1]：不操作 dp[i][j] = dp[i-1][j-1]
② word1[i-1] ！= word2[j-1]：
操作一：word1删除一个字符：dp[i][j] = dp[i-1][j] + 1
操作二：word1添加一个字符 == word2删除一个字符：dp[i][j] = dp[i][j-1] + 1
操作三：word1替换一个字符(替换成word2[j-1]):dp[i][j] = dp[i-1][j-1] + 1
3、初始化：dp[i][0] = i, dp[0][j] = j

代码

class Solution:
    # 1、含义：dp[i][j]表示以word1[i-1] word2[j-1]结尾的单词的最近编辑距离
    # 2、递推：
    #    ① word1[i-1] == word2[j-1]：不操作 dp[i][j] = dp[i-1][j-1]
    #    ② word1[i-1] ！= word2[j-1]：
    #       操作一：word1删除一个字符：dp[i][j] = dp[i-1][j] + 1
    #       操作二：word1添加一个字符 == word2删除一个字符：dp[i][j] = dp[i][j-1] + 1
    #       操作三：word1替换一个字符(替换成word2[j-1]):dp[i][j] = dp[i-1][j-1] + 1
    # 3、初始化：dp[i][0] = i, dp[0][j] = j
    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 + 1):
            dp[i][0] = i
        for j in range(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(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
        return dp[-1][-1]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23