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## Problem

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.

## Code

word1 = “horse”, word2 = “ros”

r o s
0 1 2 3
0 0 1 2 3
h 1 1 1 2 3
o 2 2 2 1 2
r 3 3 2 2 2
s 4 4 3 3 2
e 5 5 4 4 3
class Solution {
public int minDistance(String word1, String word2) {
int len1 = word1.length();
int len2 = word2.length();

int[][] dp = new int[len1 + 1][len2 + 1];
for(int j = 1; j <= len2; j++){
dp[0][j] = j;
}

for(int i = 1; i <= len1; i++){
dp[i][0] = i;
}

for(int i = 1; i <= len1; i++){
for(int j = 1; j <= len2; j++){
char c1 = word1.charAt(i - 1);
char c2 = word2.charAt(j - 1);

if(c1 == c2){
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
}
}
}

return dp[len1][len2];
}
}