【LeetCode】72. Edit Distance 解题报告（Python & C++）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/edit-distance/description/

题目描述

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

题目大意

给了两个字符串，现在有三种操作，问最少做多少次操作，能使word1变成word2。三种操作是：

1. 插入一个字符
2. 删除一个字符
3. 替换一个字符

解题方法

记忆化搜索

程序的世界真是其妙无穷。这个题和最长公共子序列非常相似，需要判断最后的一个字符是否相等。如果相等，那么只用判断前面的子串即可，如果不等，需要前面子串变成相等之后+1（替换）把word1变成word2。

图源花花酱：

所以，可以使用记忆化搜索，代码比较简单，需要注意的是初始化的数组大小是L1 + 1和L2 + 1，因为函数的意义是[0, L1], [0, L2]区间变成相等的最小操作次数，闭区间。

Python代码如下：

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        L1, L2 = len(word1), len(word2)
        dp = [[-1] * (L2 + 1) for _ in range(L1 + 1)]
        return self.getDistance(word1, word2, dp, L1, L2)
    
    def getDistance(self, word1, word2, dp, pos1, pos2):
        if pos1 == 0: return pos2
        if pos2 == 0: return pos1
        if dp[pos1][pos2] >= 0: return dp[pos1][pos2]
        
        res = 0
        if word1[pos1 - 1] == word2[pos2 - 1]:
            res = self.getDistance(word1, word2, dp, pos1 - 1, pos2 - 1)
        else:
            res = min(self.getDistance(word1, word2, dp, pos1 - 1, pos2 - 1),
                      self.getDistance(word1, word2, dp, pos1, pos2 - 1),
                      self.getDistance(word1, word2, dp, pos1 - 1, pos2)) + 1
        dp[pos1][pos2] = res
        return res

C++代码如下：

class Solution {
public:
    int minDistance(string word1, string word2) {
        const int L1 = word1.size();
        const int L2 = word2.size();
        dp_ = vector<vector<int>>(L1 + 1, vector<int>(L2 + 1, -1));
        return getDistance(word1, word2, L1, L2);
    }
private:
    vector<vector<int>> dp_;
    int getDistance(string& word1, string& word2, int l1, int l2) {
        if (l1 == 0) return l2;
        if (l2 == 0) return l1;
        if (dp_[l1][l2] >= 0) return dp_[l1][l2];
        
        int res = 0;
        if (word1[l1 - 1] == word2[l2 - 1])
            res = getDistance(word1, word2, l1 - 1, l2 - 1);
        else
            res = min(getDistance(word1, word2, l1 - 1, l2 - 1),
                     min(getDistance(word1, word2, l1 - 1, l2),
                     getDistance(word1, word2, l1, l2 - 1))) + 1;
        dp_[l1][l2] = res;
        return res;
    }
};

动态规划

知道了记忆化搜索之后，很容易改成动态规划。事实上，这个题的动态规划比记忆化搜索还要慢。

python代码如下：

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        L1, L2 = len(word1), len(word2)
        dp = [[0] * (L2 + 1) for _ in range(L1 + 1)]
        for i in range(L1 + 1):
            dp[i][0] = i
        for j in range(L2 + 1):
            dp[0][j] = j
        for i in range(1, L1 + 1):
            for j in range(1, L2 + 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
        return dp[L1][L2]

C++代码如下：

class Solution {
public:
    int minDistance(string word1, string word2) {
        const int L1 = word1.size();
        const int L2 = word2.size();
        vector<vector<int>> dp(L1 + 1, vector<int>(L2 + 1, -1));
        for (int i = 0; i <= L1; i++)
            dp[i][0] = i;
        for (int j = 0; j <= L2; j++)
            dp[0][j] = j;
        for (int i = 1; i <= L1; i++) {
            for (int j = 1; j <= L2; j++) {
                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], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[L1][L2];
    }
};

日期

2018 年 12 月 10 日 —— 又是周一！