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‘)
 

題意

求最小編輯距離

有3個操作：刪除，插入和替換

題解

一開始想了個dp，不夠快

 1 class Solution {
 2 public:
 3     int minDistance(string word1, string word2) {
 4         int l1 = word1.length(), l2 = word2.length();
 5         vector<vector<int>>dp(l1+1, vector<int>(l2+1, INT_MAX));
 6         for (int
 
 i = 0; i <= l1; i++)dp[i][0] = i;
 7         for (int i = 0; i <= l2; i++)dp[0][i] = i;
 8         for (int i = 1; i <= l1; i++) {
 9             int minnum = INT_MAX;
10             for (int j = 1; j <= l2; j++) {
11                 if (word1[i-1] == word2[j-1])
12                     dp[i][j] =  dp[i - 1][j - 1];
13                 else {
14                     int dp1 = dp[i - 1][j] + 1;
15                     int dp2 = dp[i - 1][j - 1] + 1;
16                     dp[i][j] = min(dp1, dp2);
17                     if (minnum != INT_MAX)
18                         dp[i][j] = min(minnum + j, dp[i][j]);
19                 }
20                 minnum = min(dp[i][j] - j, minnum);
21             }
22         }
23         return dp[l1][l2];
24     }
25 };

其實是有個地方我沒註意到： dp[i][j] 與 dp[i][j-1] 的關系實際上是映射著插入操作的

 1 class Solution {
 2 public:
 3     int minDistance(string word1, string word2) {
 4         int l1 = word1.length(), l2 = word2.length();
 5         vector<vector<int>>dp(l1+1, vector<int>(l2+1, INT_MAX));
 6         for (int i = 0; i <= l1; i++)dp[i][0] = i;
 7         for (int i = 0; i <= l2; i++)dp[0][i] = i;
 8         for (int i = 1; i <= l1; i++) {
 9             for (int j = 1; j <= l2; j++) {
10                 if (word1[i-1] == word2[j-1])
11                     dp[i][j] =  dp[i - 1][j - 1];
12                 else {
13                     dp[i][j] = min(dp[i - 1][j], min(dp[i - 1][j - 1], dp[i][j - 1])) + 1;
14                 }
15             }
16         }
17         return dp[l1][l2];
18     }
19 };

貌似換成數組會更快，我還是第一次知道可以不動態申請內存這樣寫數組……

 1 class Solution {
 2 public:
 3     int minDistance(string word1, string word2) {
 4         int l1 = word1.length(), l2 = word2.length();
 5         int dp[l1+1][l2+1];
 6         for (int i = 0; i <= l1; i++)dp[i][0] = i;
 7         for (int i = 0; i <= l2; i++)dp[0][i] = i;
 8         for (int i = 1; i <= l1; i++) {
 9             for (int j = 1; j <= l2; j++) {
10                 if (word1[i-1] == word2[j-1])
11                     dp[i][j] =  dp[i - 1][j - 1];
12                 else {
13                     dp[i][j] = min(dp[i - 1][j], min(dp[i - 1][j - 1], dp[i][j - 1])) + 1;
14                 }
15             }
16         }
17         return dp[l1][l2];
18     }
19 };

19.2.13 [LeetCode 72] Edit Distance