【LeetCode】72. Edit Distance

1.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

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

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')

2.Solutions

Let following be the function definition :-

f(i, j) := minimum cost (or steps) required to convert first i characters of word1 to first j characters of word2

Case 1: word1[i] == word2[j], i.e. the ith the jth character matches.

f(i, j) = f(i - 1, j - 1)

Case 2: word1[i] != word2[j], then we must either insert, delete or replace, whichever is cheaper

f(i, j) = 1 + min { f(i, j - 1), f(i - 1, j), f(i - 1, j - 1) }

1. f(i, j - 1) represents insert operation
2. f(i - 1, j) represents delete operation
3. f(i - 1, j - 1) represents replace operation

Here, we consider any operation from word1 to word2. It means, when we say insert operation, we insert a new character after word1 that matches the jth character of word2. So, now have to match i characters of word1 to j - 1 characters of word2. Same goes for other 2 operations as well.

Note that the problem is symmetric. The insert operation in one direction (i.e. from word1 to word2) is same as delete operation in other. So, we could choose any direction.

Above equations become the recursive definitions for DP.

Base Case:

f(0, k) = f(k, 0) = k

Below is the direct bottom-up translation of this recurrent relation. It is only important to take care of 0-based index with actual code :-

public int minDistance(String word1, String word2) {
    int m = word1.length();
    int n = word2.length();

    int[][] cost = new int[m + 1][n + 1];
    for(int i = 0; i <= m; i++)
        cost[i][0] = i;
    for(int i = 1; i <= n; i++)
        cost[0][i] = i;

    for(int i = 0; i < m; i++) {
        for(int j = 0; j < n; j++) {
            if(word1.charAt(i) == word2.charAt(j))
                cost[i + 1][j + 1] = cost[i][j];
            else {
                int a = cost[i][j];
                int b = cost[i][j + 1];
                int c = cost[i + 1][j];
                cost[i + 1][j + 1] = a < b ? (a < c ? a : c) : (b < c ? b : c);
                cost[i + 1][j + 1]++;
            }
        }
    }
    return cost[m][n];
}

补DP基础：

Python - 3000字长文解释DP基础 | 公瑾™

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