Description Link to heading

72.edit-distance

Solution Link to heading

It’s easy to consider what dp[i][j] should denotes.

  • if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
  • else, we can consider in three cases:
    • replace the word1[i - 1]: dp[i][j] = dp[i - 1][j - 1] + 1;
    • remove word1[i - 1]: dp[i][j] = dp[i - 1][j] + 1;
    • insert word2[j - 1] between word[i - 1] and word[i], it’s the same as remove word2[j - 1]: dp[i][j] = dp[i][j - 1] + 1;

We should also pay attention to the initialzation of dp[i][j].

Code Link to heading 

class Solution {
  public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
        for (int i = 1; i <= word1.size(); i++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= word2.size(); j++) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= word1.size(); i++) {
            for (int j = 1; j <= word2.size(); j++) {
                if (word1[i - 1] == word2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1];
                else {
                    dp[i][j] = min(min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};