# 72. Edit Distance

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.

``````class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size();
vector<vector<int>> memo(m, vector<int>(n));
return dfs(word1, 0, word2, 0, memo);
}
int dfs(string& word1, int i, string& word2, int j, vector<vector<int>>& memo) {
int m = word1.size(), n = word2.size(), res = 0;
if (i == m) return n - j;
if (j == n) return m - i;
if (memo[i][j] > 0) return memo[i][j];
if (word1[i] == word2[j]) {
return dfs(word1, i + 1, word2, j + 1, memo);
} else {
int insertCnt = dfs(word1, i, word2, j + 1, memo);
int deleteCnt = dfs(word1, i + 1, word2, j, memo);
int replaceCnt = dfs(word1, i + 1, word2, j + 1, memo);
res = min(insertCnt, min(deleteCnt, replaceCnt)) + 1;
}
return memo[i][j] = res;
}
};
``````

``````  Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2
``````

dp[i][j] = / dp[i - 1][j - 1] if word1[i - 1] == word2[j - 1]

\ min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1 else

``````class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
for (int i = 0; i <= m; ++i) dp[i][0] = i;
for (int i = 0; i <= n; ++i) dp[0][i] = i;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++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[m][n];
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/72

One Edit Distance

Delete Operation for Two Strings

Minimum ASCII Delete Sum for Two Strings

Uncrossed Lines

Minimum White Tiles After Covering With Carpets

https://leetcode.com/problems/edit-distance/discuss/25846/C%2B%2B-O(n)-space-DP

LeetCode All in One 题目讲解汇总(持续更新中…)

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