LeetCode: 72. Edit Distance

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本文深入解析编辑距离算法，一种用于衡量两个字符串相似度的动态规划方法。通过具体实例，阐述了如何通过插入、删除和替换操作，计算从一个字符串转换为另一个字符串所需的最小操作数。文章提供了详细的算法步骤和C++实现代码。

72. Edit Distance

题目
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:

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')

思路解析
- 这道题是一道动态规划的题，需要将一个字符串word1经过add、delete、replace三种操作变换成另一个字符串word2，求最小的编辑距离.
- 解题思路：假设word1=[123...n] ,word2=[123...m]，要求使得word1转换成word2。对于这个问题，我们将其分解为子问题求解。
  - 定义dis[i][j]
    - 表示word1' = [1..i]转换成word2' = [1...j]的编辑距离（i代表word1前i个字符,j代表word2前j个字符）
    - 因此word1到word2的编辑距离为dis[n][m]
  - 求解word1到word2的编辑距离，我们可以求取word1的前i个字符(0 < i < n)到word2的前j个字符(0 < j < m)的编辑距离dis[i][j]。当然每个dis[i][j]都基于之前的计算。
  - 步骤
    - 初始化
      - dis[i, 0] = i
      - dis[0, j] = j
    - 递推关系–核心
      
      $\begin{cases} dis[i][j-1] + 1\\ dis[i-1][j] + 1\\ dis[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1))\\ \end{cases}$
      其中三个操作的表示
      - insert: dis[i, j] = dis[i][j-1] + 1
      - delete: dis[i, j] = dis[i-1][j] + 1
      - replace or no op：dis[i, j] = dis[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1)
      对于每dis[i][j]，我们选取最小编辑距离
  - 最后得到的dis[n][m]就是word1到word2的编辑距离

实现代码

class Solution {
public:
    int minDistance(string word1, string word2) {
		int m = word1.size();
		int n = word2.size();
		int dis[m+1][n+1] = {0};

		for (int i = 0; i <= m; ++i)
			dis[i][0] = i;

		for (int j = 0;j <= n; ++j)
			dis[0][j] = j;

		// m or n equal 0
		if (!m && !n)
			return max(m, n);

		// insert: 	dis[i, j] = dis[i][j-1] + 1	
		// delete:	dis[i, j] = dis[i-1][j] + 1
		// replace or no op：	dis[i, j] = dis[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1)

		for (int i = 1; i <= m; ++i) {			
			for (int j = 1; j <= n; ++j) {
				dis[i][j] = min(dis[i][j-1] + 1,
					min(dis[i-1][j] + 1, dis[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1)));
			}
		}
		return dis[m][n];
    }
};

参考链接 - pdf