Edit Distance

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本文介绍了一种使用动态规划解决两个字符串之间的编辑距离问题的方法。通过插入、删除和替换字符的操作，实现从一个字符串转换为另一个字符串所需的最少步骤。文章提供了一个C++实现示例，展示了如何初始化和更新动态规划表格。

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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

解题思路：

又是一道水DP

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int dp[1010][1010];
        memset(dp, 0x3f3f3f3f, sizeof(dp));
        dp[0][0] = 0;
        for(int i = 1; i <= len1; ++i) {
            dp[i][0] = i;
        }
        for(int i = 1; i <= len2; ++i) {
            dp[0][i] = i;
        }
        for(int i = 1; i <= len1; ++i) {
            for(int j = 1; j <= len2; ++j) {
                if(word1[i-1] == word2[j-1]) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = min(dp[i][j], dp[i-1][j-1] + 1);
                    dp[i][j] = min(dp[i][j], dp[i][j-1] + 1);
                    dp[i][j] = min(dp[i][j], dp[i-1][j] + 1);
                }
            }
        }
        return dp[len1][len2];
    }
};