Leetcode: Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

分析：用动态规划来解，f[i][j]表示word1[0,i-1]与word2[0,j-1]的edit distance。当word1[i] == word2[j]时，f[i][j] = f[i-1][j-1]；否则，f[i][j]为f[i-1][j]+1（delete a character）、f[i][j-1]+1（insert a character）、f[i-1][j-1]+1（replace a character）中的最小值。

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.length(), n = word2.length();
        if(m == 0 || n == 0) return abs(m-n);
        int record[m+1][n+1];
        for(int i = 0; i <= m; i++)
            record[i][0] = i;
        for(int j = 0; j <= n; j++)
            record[0][j] = j;
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++){
                if(word1[i-1] == word2[j-1]) record[i][j] = record[i-1][j-1];
                else{
                    int min_step = min(record[i-1][j],record[i][j-1]);
                    record[i][j] = min(min_step,record[i-1][j-1]) + 1;
                }
            }
        return record[m][n];
    }
};