題目鏈接：https://leetcode-cn.com/problems/edit-distance/
描述：給定兩個(gè)單詞 word1 和 word2，計(jì)算出將 word1 轉(zhuǎn)換成 word2 所使用的最少操作數(shù) 。
你可以對(duì)一個(gè)單詞進(jìn)行如下三種操作：

插入一個(gè)字符
刪除一個(gè)字符
替換一個(gè)字符

示例 1:

輸入: word1 = "horse", word2 = "ros"
輸出: 3
解釋: 
horse -> rorse (將 'h' 替換為 'r')
rorse -> rose (刪除 'r')
rose -> ros (刪除 'e')

示例 2:

輸入: word1 = "intention", word2 = "execution"
輸出: 5
解釋: 
intention -> inention (刪除 't')
inention -> enention (將 'i' 替換為 'e')
enention -> exention (將 'n' 替換為 'x')
exention -> exection (將 'n' 替換為 'c')
exection -> execution (插入 'u')

• 解法一：傳統(tǒng)的遞歸形式（會(huì)有最大遞歸深度的限制，超時(shí)）

# 遞歸的方法，但是時(shí)間復(fù)雜度較高，需要使用一個(gè)dp數(shù)組記錄重復(fù)操作的步驟
    def minDistance(word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        if word1 == '':
            return len(word2)
        if word2 == '':
            return len(word1)

        if word1[-1] == word2[-1]:
            return self.minDistance2(word1[:-1], word2[:-1])
        else:
            res = 1 + min(
                self.minDistance2(word1[:-1], word2[:-1]),
                self.minDistance2(word1, word2[:-1]),
                self.minDistance2(word1[:-1], word2)
            )
        return res

• 解法二：動(dòng)態(tài)規(guī)劃，相當(dāng)于空間換時(shí)間

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        n = len(word1)
        m = len(word2)
        dp = [[0 for j in range(m+1)] for i in range(n+1)]
        for i in range(n+1):
            for j in range(m+1):
                if i == 0:
                    dp[i][j] = j
                    continue
                if j == 0:
                    dp[i][j] = i
                    continue
                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
        return dp[n][m]