72. Edit Distance

Problem:

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')

思路：

Solution 1.1 (C++):

int minDistance(string word1, string word2) {
    int m = word1.size(), n = word2.size();
    vector> dp(m+1, vector(n+1, 0));
    
    for (int i = 1; i <= m; ++i)  dp[i][0] = i;
    for (int j = 1; j <= n; ++j)  dp[0][j] = j;
    
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++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[m][n];
}

性能：

Runtime: 28 ms  Memory Usage: 11.2 MB

Solution 1.2 (C++):

int minDistance(string word1, string word2) {
    int m = word1.size(), n = word2.size();
    vector pre(n+1, 0), cur(n+1, 0);
    
    for (int j = 1; j <= n; ++j)  pre[j] = j;
    
    for (int i = 1; i <= m; ++i) {
        cur[0] = i;
        for (int j = 1; j <= n; ++j) {
            if (word1[i-1] == word2[j-1])  cur[j] = pre[j-1];
            else
                cur[j] = min(pre[j-1], min(cur[j-1], pre[j])) + 1;
                //dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;
        }
        fill(pre.begin(), pre.end(), 0);
        swap(pre, cur);
    }
    return pre[n];
}

性能：

Runtime: 8 ms  Memory Usage: 8.7 MB