算法训练Day56： 583. 两个字符串的删除操作 72. 编辑距离

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两个字符串的删除操作

Category	Difficulty	Likes	Dislikes	ContestSlug	ProblemIndex	Score
algorithms	Medium (66.47%)	578	0	-	-	0

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Companies

给定两个单词 word1 和 word2 ，返回使得 word1 和 word2 相同所需的最小步数。

每步 可以删除任意一个字符串中的一个字符。

示例 1：

输入: word1 = "sea", word2 = "eat"
输出: 2
解释: 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea"

示例 2:

输入：word1 = "leetcode", word2 = "etco"
输出：4

提示：

• 1 <= word1.length, word2.length <= 500
• word1 和 word2 只包含小写英文字母

---

Discussion | Solution

题解

// @lc code=start
class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size() + 1, vector(word2.size() + 1));
        for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
        for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
        for (int i = 1; i <= word1.size(); i++) {
            for (int j = 1; j <= word2.size(); 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, dp[i][j - 1] + 1);
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

编辑距离

Category	Difficulty	Likes	Dislikes	ContestSlug	ProblemIndex	Score
algorithms	Hard (62.79%)	2958	0	-	-	0

Tags

Companies

给你两个单词 word1 和 word2， 请返回将 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：

• 插入一个字符
• 删除一个字符
• 替换一个字符

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')

提示：

• 0 <= word1.length, word2.length <= 500
• word1 和 word2 由小写英文字母组成

---

Discussion | Solution

题解

// @lc code=start
class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size() + 1, vector(word2.size() + 1, 0));
        for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
        for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
        for (int i = 1; i <= word1.size(); i++) {
            for (int j = 1; j <= word2.size(); 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], dp[i - 1][j], dp[i][j - 1]}) + 1;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};