583. 兩個字符串的刪除操作

• 思路
  • example
  • 最小步數
  • 雙串DP

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

• 復雜度. 時間：O(mn), 空間: 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. 編輯距離

• 思路
  • example
  • 最少操作數
    • 插入
    • 刪除

    • 替換

      
      

dp[i][j]: 將word1[0,...,i-1] 轉換成word2[0,...,j-1]所使用的最少操作數。

if (word1[i - 1] == word2[j - 1])
    不操作
if (word1[i - 1] != word2[j - 1])
    增
    刪
    換
# 其實刪和增可以“轉換”，用一個case統一即可。

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
                    #              替換           插入／刪除統一處理：有兩種情況
        return dp[m][n]

• 如果需要保存最佳操作

  
  

編輯距離問題 小結

• TBA

code