LeetCode: 72. Edit Distance(Week 9)

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

        $dis[i][j] = min \begin{cases} dis[i][j-1] + 1\\ dis[i-1][j] + 1\\ dis[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1))\\ \end{cases}$
        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