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 + 1];
    for (int i = 0; i <= wlen1; i++)
        dp[i] = new int[wlen2 + 1];
    
    //int dp[maxn][maxn] = { 0 };
    for (int i = 0; i <= wlen1; i++)
        dp[i][0] = i;
    for (int j = 0; j <= wlen2; j++)
        dp[0][j] = j;
    int temp = 0;
    for (int i = 1; i <= wlen1; i++){
        for (int j = 1; j <= wlen2; j++){
            if (word1[i - 1] == word2[j - 1])dp[i][j] = dp[i - 1][j-1];
            else{
                temp = dp[i - 1][j - 1]<dp[i - 1][j] ? dp[i - 1][j - 1] : dp[i - 1][j];
                temp = temp < dp[i][j - 1] ? temp : dp[i][j - 1];
                dp[i][j] = temp + 1;
            }
        }
    }

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

    return dp[wlen1][wlen2];
}