歡迎討論～

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

二維動態規劃dp[][]定義：dp[i][j]: word1 的前i個字元到word2前j個字元的distance。

通項公式：

example 1: Insert a character

dp[5][2]: horse->rorse->rose->ros->ro

dp[5][3]: (horse->)horse's'->rorse's'->rose's'->ros's'->ro's'

example 2: Delete a character

dp[4][3]: hors->rors->ros

dp[5][3]: hors'e'->rors'e'->ros'e'(->ros)

example 3: Replace a character

dp[4][2]: hors->rors->ros->ro

replace word1[5] with word2[3]("s") when word1[5]!=word2[3].

dp[5][3]: hors'e'->rors'e'->ros'e'->ro's'

因此，如果要計算得出dp[i][j] 可以通過{dp[i-1][j-1],dp[i][j-1],dp[i-1][j]} 加上一種基礎操作的到。

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