30 Day Challenge Day 22 | Leetcode 72. Edit Distance

题解

Hard

动态规划

这道题是一道很具有代表性的二维动态规划问题，非常具有示范效应，理解了这道题的思路，能够举一反三。

一开始没有思路的时候，用一个小例子手动演算一下很有帮助。

    Example: word1: abc, word2: abbc
           \0 a b b c
        \0  0 1 2 3 4
         a  1 0 1 2 3 
         b  2 1 0 1 2
         c  3 2 1 1 1
         
    DP:
        state: 
            dp[i][j] word1[0-i] --> word2[0-j]
        initialization:
            1st row: dp[0][j] = j;
            1st col: dp[i][0] = i;
        transition:
            case: word1[i-1] == word2[j-1] ==> dp[i][j] = dp[i-1][j-1]
            case: word1[i-1] != word2[j-1] ==> dp[i][j] = min(dp[i][j-1], dp[i-1][j]) + 1  <== missing dp[i-1][j-1]
                                                                delete      insert                       replace
        result:
            dp[m][n]
    *****/

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));
        
        // initialization

        for(int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        
        for(int j = 0; j <= n; j++) {
            dp[0][j] = j;   
        }

        // state transition
        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];
    }
};