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 len1 = word1.length();
int len2 = word2.length();
int[][] dp = new int[len1+1][len2+1];
for(int i=0;i<=len1;i++){
dp[i][0] = i;
}
for(int j=0;j<=len2;j++){
dp[0][j] = j;
}
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
if(word1.charAt(i - 1) == word2.charAt(j - 1)){
// 两个字符完全相同
dp[i][j] = dp[i - 1][j - 1];
}else{
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]))+1;
}
}
}
return dp[len1][len2];
}
}