# 72.Edit-Distance

## 题目描述

``````Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 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')``````

## 思路

``````state: f[i][j] word1的前i个字符最少要用几次编辑可以变成word2的前j个字符

function: f[i][j]
= MIN(f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1]) // word1[i] == word2[j]
= MIN(f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1] + 1) //
word1[i] != word2[j]

initialize: f[i][0] = i, f[0][j] = j

insert: f[i][j] = f[i][j-1] + 1
remove: f[i][j] = f[i-1][j] + 1
replace: f[i][j] = f[i-1][j-1] + 1

if (word1.charAt(i-1) == word2.charAt(j-1)){
dp[i][j] = dp[i-1][j-1];
} else {
dp[i][j] = dp[i-1][j-1] + 1;   //replace operation
}
dp[i][j] = Math.min(dp[i][j], Math.min(dp[i][j-1]+1, dp[i-1][j]+1));``````

## 代码

### Approach #1 Dynamic Programming

• `如果 s1[i-1]和s2[j]可以配对，那我就删除s1[i]即可(删除)`

• `如果 s1[i]和s2[j-1]可以配对，那我就在s1的后面加上s2[j]即可(插入)`

• `如果 s1[i-1]和s2[j-1]可以配对，那我就把s1[i]修改成s2[j]即可`

``````public class Solution {
public int minDistance(String word1, String word2) {
char[] s1 = word1.toCharArray();
char[] s2 = word2.toCharArray();
int i, j;
int m = s1.length;
int n = s2.length;

int[][] f = new int[m + 1][n + 1];

for (i = 0; i <= m; ++i) {
for (j = 0; j <= n; ++j) {
if (i == 0) {
f[i][j] = j;
continue;
}

if (j == 0) {
f[i][j] = i;
continue;
}

f[i][j] = Math.min(Math.min(f[i - 1][j], f[i][j - 1]), f[i - 1][j - 1]) + 1;

if (s1[i - 1] == s2[j - 1]) {
f[i][j] = Math.min(f[i][j], f[i - 1][j - 1]);
}

}
}

return f[m][n];
}
}``````

### Approach #2

``````public class Solution {
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();

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

for (int i = 1; i < n + 1; i++) {
for (int j = 1; j < m + 1; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
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[n][m];
}
}``````

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