LeetCode每日一题(72. Edit Distance)-优快云博客

本文探讨了如何解决LeetCode上的第72题——编辑距离问题，介绍了如何通过插入、删除和替换字符将一个字符串转换为另一个字符串，并提供了两个示例解释了操作过程。同时，文章详细阐述了实现该算法的思路和约束条件。

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Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

Insert a character
Delete a character
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’)

Constraints:

0 <= word1.length, word2.length <= 500
word1 and word2 consist of lowercase English letters.

对于 word1[i]和 word2[j]，我们能做的操作就是题目里给的那三种:

插入，在 word1[i]处插入 word2[j], 此时 j += 1, 而 i 不变, 操作计数+1
删除, 把 word1[i]删除, 此时 i += 1, 而 j 不变, 操作计数+1
替换, 把 word1[i]替换成 word2[j], 此时 i += 1, j += 1, 操作计数视 word1[i]和 word2[j]而定，如果 word1[i] == word2[j], 则说明实际不用替换，操作计数不变，否则操作计数+1

最终，如果 i 先到达 word1 尾部，则证明我们需要往 word1 尾部插入 word2 的剩余字符，如果 j 先到达 word2 的尾部，则证明我们需要删除 word1 的剩余字符


use std::collections::HashMap;

impl Solution {
    fn dp(chars1: &Vec<char>, chars2: &Vec<char>, i: usize, j: usize, cache: &mut HashMap<(usize, usize), i32>) -> i32 {
        if j == chars2.len() {
            return (chars1.len() - i) as i32;
        }
        if i == chars1.len() {
            return (chars2.len() - j) as i32;
        }
        if let Some(c) = cache.get(&(i, j)) {
            return *c;
        }
        let insert = Solution::dp(chars1, chars2, i, j + 1, cache) + 1;
        let delete = Solution::dp(chars1, chars2, i + 1, j, cache) + 1;
        let replace = Solution::dp(chars1, chars2, i + 1, j + 1, cache) + if chars1[i] == chars2[j] { 0 } else { 1 };
        let ans = insert.min(delete).min(replace);
        cache.insert((i, j), ans);
        ans
    }
    pub fn min_distance(word1: String, word2: String) -> i32 {
        Solution::dp(&word1.chars().collect(), &word2.chars().collect(), 0, 0, &mut HashMap::new())
    }
}