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

 

 int minDistance(string word1,string word2){
     int wlen1 = word1.size();
     int wlen2 = word2.size();

     int**dp = new int*[wlen1 + ];
     for (int i = ; i <= wlen1; i++)
         dp[i] = new int[wlen2 + ];

     //int dp[maxn][maxn] = { 0 };
     for (int i = ; i <= wlen1; i++)
         dp[i][] = i;
     for (int j = ; j <= wlen2; j++)
         dp[][j] = j;
     int temp = ;
     for (int i = ; i <= wlen1; i++){
         for (int j = ; j <= wlen2; j++){
             if (word1[i - ] == word2[j - ])dp[i][j] = dp[i - ][j-];
             else{
                 temp = dp[i - ][j - ]<dp[i - ][j] ? dp[i - ][j - ] : dp[i - ][j];
                 temp = temp < dp[i][j - ] ? temp : dp[i][j - ];
                 dp[i][j] = temp + ;
             }
         }
     }

     /*
     for (int i = 0; i <= wlen1; i++)
         delete[]dp[i];
     delete[]dp;
     */

     return dp[wlen1][wlen2];
 }