动态规划——Edit Distance

大意：给定两个字符串word1和word2，为了使word1变为word2，可以进行增加、删除、替换字符三种操作，请输出操作的最少次数

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‘)

状态：dp[i][j]把word1[0..i-1]转换到word2[0..j-1]的最少操作次数
状态转移方程：

　　(1)如果word1[i-1] == word2[j-1]，则令dp[i][j] = dp[i-1][j-1]
　　(2)如果word1[i-1] != word2[j-1]，由于没有一个特别有规律的方法来断定执行何种操作，在增加、删除、替换三种操作中选一种操作次数少的赋值给dp[i][j];
　　　 增加操作：dp[i][j] = dp[i][j-1] + 1
　　　 删除操作：dp[i][j] = dp[i-1][j] + 1

　　     替换操作：dp[i][j] = dp[i-1][j-1] + 1

 1 int minDistance(string word1,string word2){
 2     int wlen1 = word1.size();
 3     int wlen2 = word2.size();
 4
 5     int**dp = new int*[wlen1 + 1];
 6     for (int i = 0; i <= wlen1; i++)
 7         dp[i] = new int[wlen2 + 1];
 8
 9     //int dp[maxn][maxn] = { 0 };
10     for (int i = 0; i <= wlen1; i++)
11         dp[i][0] = i;
12     for (int j = 0; j <= wlen2; j++)
13         dp[0][j] = j;
14     int temp = 0;
15     for (int i = 1; i <= wlen1; i++){
16         for (int j = 1; j <= wlen2; j++){
17             if (word1[i - 1] == word2[j - 1])dp[i][j] = dp[i - 1][j-1];
18             else{
19                 temp = dp[i - 1][j - 1]<dp[i - 1][j] ? dp[i - 1][j - 1] : dp[i - 1][j];
20                 temp = temp < dp[i][j - 1] ? temp : dp[i][j - 1];
21                 dp[i][j] = temp + 1;
22             }
23         }
24     }
25
26     /*
27     for (int i = 0; i <= wlen1; i++)
28         delete[]dp[i];
29     delete[]dp;
30     */
31
32     return dp[wlen1][wlen2];
33 }

原文地址：https://www.cnblogs.com/messi2017/p/9892109.html