题目描述

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

样例

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')

算法1

DP $O(m x n)$

C++ 代码

    // d[i][j] : w1的前i个char convert to w2的前j个char 的distance

    // 为了 索引 word 方便， w1 = " " + w1, w2 = " " + w2;

    // if w1[i] == w2[j]: d[i][j] = 0;
    // else:  d[i][j] = 1 + min { d[i][j-1], d[i-1][j], d[i-1][j-1] }
    // init: d[i][0] = i, d[0][j] = j

    int minDistance(string word1, string word2) {
        string w1 = " " + word1, w2 = " " + word2;
        int m = w1.size(), n = w2.size();

        vector<vector<int>> d(m, vector<int>(n, INT_MAX));

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

        for(int i=1; i<m; i++) {
            for(int j=1; j<n; j++) {
                if(w1[i] == w2[j]) { d[i][j] = d[i-1][j-1]; }
                else { d[i][j] = 1+ min( min( d[i][j-1], d[i-1][j] ), d[i-1][j-1] ); }
            }
        }

        return d[m-1][n-1];
    }