[leetcode]编辑距离——python版本

Leetcode之编辑距离

题目：

定义dp函数：dp(i,j)表示word1[0…i]转成word2[0…j]的最少操作数

def minDistance(word1,word2):
    def dp(i,j):
        #递归的终点
        #word1走到头，要将word2剩余的字符插入
        if i == -1:
            return j + 1
        #word2走到头，要将word1剩余的字符删除
        if j == -1:
            return i + 1
        #相同则不动，指针向前移动
        if word1[i] == word2[j]:
            return dp(i-1,j-1)
        else:
            return min(
            	dp(i,j-1) + 1,#插入
                dp(i-1,j) + 1,#删除
                dp(i-1,j-1) + 1#替换
            )
    return dp(len(word1)-1,len(word2)-1)

但该方法超出时间限制，需要用动态规划优化一下

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

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        def dp(i,j):
            if i < 0:
                return j + 1
            if j < 0:
                return i + 1
            if (i,j) in memo:
                return memo[(i,j)]
            if word1[i] == word2[j]:
                memo[(i,j)] = dp(i-1,j-1)
            else:
                memo[(i,j)] = min(
                    dp(i-1,j) + 1,
                    dp(i,j-1) + 1,
                    dp(i-1,j-1) + 1
                )
            return memo[(i,j)]
        m = len(word1)
        n = len(word2)
        memo = dict()
        return dp(m-1,n-1)

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m = len(word1)
        n = len(word2)
        #dp[i][j]表示word1[0..i]转换成word2[0..j]最少操作数，由于base case的index为-1，故整体右移一位
        dp = [[0 for i in range(n+1)] for j 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] + 1,
                        dp[i][j-1] + 1,
                        dp[i-1][j-1] + 1
                    )
        return dp[m][n]