Day 56 DP：583. 两个字符串的删除操作, 72. 编辑距离

583. 两个字符串的删除操作

思路
- example
- 最小步数
- 双串DP

dp[i][j]: word1 前i个，word2前j个（word1: 0, ..., i-1; word2: 0, ..., j-1)

复杂度. 时间：O(mn), 空间: O(mn)

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

72. 编辑距离

思路
- example
- 最少操作数
  - 插入
  - 删除
  - 替换

dp[i][j]: 将word1[0,...,i-1] 转换成word2[0,...,j-1]所使用的最少操作数。

if (word1[i - 1] == word2[j - 1])
    不操作
if (word1[i - 1] != word2[j - 1])
    增
    删
    换
# 其实删和增可以“转换”，用一个case统一即可。

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)  
        dp = [[0 for _ in range(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-1], dp[i-1][j], dp[i][j-1]) + 1
                    #              替换           插入／删除统一处理：有两种情况
        return dp[m][n]

如果需要保存最佳操作

编辑距离问题小结

code