## Edit Distance

### 描述

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

### 分析

1. 如果c==d，则f[i][j]=f[i-1][j-1]
2. 如果c!=d

1. 如果将c替换成d，则f[i][j]=f[i-1][j-1]+1
2. 如果在c后面添加一个d，则f[i][j]=f[i][j-1]+1
3. 如果将c删除，则f[i][j]=f[i-1][j]+1

### 动规

// Edit Distance
// 二维动规，时间复杂度O(n*m)，空间复杂度O(n*m)
public class Solution {
public int minDistance(String word1, String word2) {
final int n = word1.length();
final int m = word2.length();
// 长度为n的字符串，有n+1个隔板
int[][] f = new int[n+1][m+1];
for (int i = 0; i <= n; i++)
f[i][0] = i;
for (int j = 0; j <= m; j++)
f[0][j] = j;

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1))
f[i][j] = f[i - 1][j - 1];
else {
int mn = Math.min(f[i - 1][j], f[i][j - 1]);
f[i][j] = 1 + Math.min(f[i - 1][j - 1], mn);
}
}
}
return f[n][m];
}
}


### 动规+滚动数组

// Edit Distance
// 二维动规+滚动数组
// 时间复杂度O(n*m)，空间复杂度O(n)
public class Solution {
public int minDistance(String word1, String word2) {
if (word1.length() < word2.length())
return minDistance(word2, word1);

int[] f = new int[word2.length() + 1];
int upper_left = 0; // 额外用一个变量记录f[i-1][j-1]

for (int i = 0; i <= word2.length(); ++i)
f[i] = i;

for (int i = 1; i <= word1.length(); ++i) {
upper_left = f[0];
f[0] = i;

for (int j = 1; j <= word2.length(); ++j) {
int upper = f[j];

if (word1.charAt(i - 1) == word2.charAt(j - 1))
f[j] = upper_left;
else
f[j] = 1 + Math.min(upper_left, Math.min(f[j], f[j - 1]));

upper_left = upper;
}
}

return f[word2.length()];
}
}