LeetCode 72. Edit Distance

72. 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

这是一道很经典的动态规划问题。使用dp[i][j]来表示word1的前i个字符转变为word2的前j个字符的编辑距离。为了方便，以下假设word1/word2的下标由1开始，即word1[1:3]表示字符串{word1[0], word[1], word[2]}。那么，dp[i][j]表示word1[1:i]和word2[1:j]的编辑距离。

假设现在已知dp[i-1][j-1]，即已经知道word1[1:i-1]到word2[1:j-1]的编辑距离，那么对于word1[i]和word2[j]，有两种情况：
1. word1[i] == wprd2[j]。此时，显然有dp[i][j] = dp[i-1][j-1]
2. word1[i] != word2[j]。这时，就有可能发生三种操作：insert, delete和replace。

考虑word1[i] != word2[j]时，若：
1. word1[1:i]通过insert可以得到word2[1:j]，即word1[1:i] == word2[1:j-1]，那么有：dp[i][j] = dp[i][j-1]+1。
2. word1[1:i]通过delete可以得到word2[1:j]，即word1[1:i-1] == word2[1:j]，那么有：dp[i][j] = dp[i-1][j]+1。
3. word1[1:i]直接通过replace可以得到word2[1:j]，即word1[1:i-1] == word2[1:j-1]，那么有：dp[i][j] = dp[i-1][j-1]+1。

综上，可以得到状态转移方程：

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], dp[i - 1][j], dp[i][j - 1]) + 1;

另外，再考虑边界条件，即：

dp[i][0] = i;
dp[0][j] = j;

代码实现：

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