[LeetCode] 72. 编辑距离

题目链接 : https://leetcode-cn.com/problems/edit-distance/

题目描述:

给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

1. 插入一个字符
2. 删除一个字符
3. 替换一个字符

示例:

示例 1:

输入: word1 = "horse", word2 = "ros"
输出: 3
解释: 
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')

示例 2:

输入: word1 = "intention", word2 = "execution"
输出: 5
解释: 
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')

思路:

动态规划:

dp[i][j]代表word1到i位置转换成word2到j位置需要最少步数

所以,

当word1[i] == word2[j],dp[i][j] = dp[i-1][j-1];

当word1[i] != word2[j],dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1

其中,dp[i-1][j-1]表示替换操作,dp[i-1][j]表示删除操作, dp[i][j-1]表示插入操作.

注意,针对第一行,第一列要单独考虑,我们引入''下图所示:

第一行,是word1为空变成word2最少步数,就是插入操作

第一列,是word2为空,需要的最少步数,就是删除操作

再附上自顶向下的方法,大家可以提供 Java 版吗?

代码:

自底向上

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n1 = len(word1)
        n2 = len(word2)
        dp = [[0] * (n2 + 1) for _ in range(n1 + 1)]
        # 第一行
        for j in range(1, n2 + 1):
            dp[0][j] = dp[0][j-1] + 1
        # 第一列
        for i in range(1, n1 + 1):
            dp[i][0] = dp[i-1][0] + 1
        for i in range(1, n1 + 1):
            for j in range(1, n2 + 1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1] ) + 1
        #print(dp)      
        return dp[-1][-1]

java

class Solution {
    public int minDistance(String word1, String word2) {
        int n1 = word1.length();
        int n2 = word2.length();
        int[][] dp = new int[n1 + 1][n2 + 1];
        // 第一行
        for (int j = 1; j <= n2; j++) dp[0][j] = dp[0][j - 1] + 1;
        // 第一列
        for (int i = 1; i <= n1; i++) dp[i][0] = dp[i - 1][0] + 1;

        for (int i = 1; i <= n1; i++) {
            for (int j = 1; j <= n2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;
            }
        }
        return dp[n1][n2];  
    }
}

自顶向下

import functools
class Solution:
    @functools.lru_cache(None)
    def minDistance(self, word1: str, word2: str) -> int:
        if not word1 or not word2:
            return len(word1) + len(word2)
        if word1[0] == word2[0]:
            return self.minDistance(word1[1:], word2[1:])
        else:
            inserted = 1 + self.minDistance(word1, word2[1:])
            deleted = 1 + self.minDistance(word1[1:], word2)
            replace = 1 + self.minDistance(word1[1:], word2[1:])
            return min(inserted, deleted, replace)