[LeetCode] 72. Edit Distance

[frdmu]

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

Insert a character
Delete a character
Replace a character
 

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')

Constraints:

• 0 <= word1.length, word2.length <= 500
• word1 and word2 consist of lowercase English letters.

 
 
 
解法：
动态规划。自己想也想不出来，直接看题解吧，看图写代码。

代码如下：

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));
        
        for (int i = 0; i < m+1; i++) {
            dp[i][0] = i;
        }
        for (int i = 0; i < n+1; i++) {
            dp[0][i] = i;
        }
        for (int i = 1; i < m+1; i++) {
            for (int j = 1; j < n+1; j++) {
                if (word1[i-1] != word2[j-1]) {
                    dp[i][j] = min(dp[i-1][j], min(dp[i-1][j-1], dp[i][j-1])) + 1;            
                } else {
                    dp[i][j] = dp[i-1][j-1];
                }
            }
        }
        
        return dp[m][n];
    }
};

Refer:
72. Edit Distance

 
又做一遍：

• 对于两个字符串的动态规划，一般思路是：使用两个指针，分别从尾部向前扫描。
• 这道题的dp table如下图所示：
  
  Refer:
  编辑距离