# LeetCode72. Edit Distance（动态规划）

## LeetCode72. Edit Distance（动态规划）

• Hard
• Accepted：167,332
• Submissions：450,132

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:

1. Insert a character
2. Delete a character
3. 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')

### 链接

https://leetcode.com/problems/edit-distance

### 题解

  f("abbc", "acc")
= f("abb", "ac")  // c = c
= 1 + min(f("ab", "ac")  // delete b
=         f("abb", "c")  // insert c
=         f("ab",  "a")) // modify b->c

$dp[i][j]=\left\{\begin{array}{cc}{dp[i-1][j-1]} & {word1[i-1] = word2[j-1]} \\ {\min (dp[i-1][j-1],dp[i][j-1],dp[i-1][j]) + 1} & {word1[i-1] \neq word2[j-1]} \end{array}\right.$

dp[i-1][j-1]对应着修改操作，dp[i][j-1]对应着插入操作，dp[i-1][j]对应着删除操作。

### 代码

• Runtime: 16 ms, faster than 77.35% of C++ online submissions for Edit Distance.
• Memory Usage: 11.3 MB, less than 23.70% of C++ online submissions for Edit Distance.
class Solution {
public:
int minDistance(string word1, string word2) {
int n = word1.length(), m = word2.length();
vector <vector<int> > dp(n + 1, vector<int>(m + 1, 0));
for (int j = 0; j <= m; ++j)
dp[0][j] = j;
for (int i = 0; i <= n; ++i)
dp[i][0] = i;
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= m; ++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[n][m];
}
};

The end.
2019年4月21日 星期日