LeetCode 72. Edit Distance (编辑距离)

原题

• 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:

  1. Insert a character
  2. Delete a character
  3. 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')
  

Reference Answer

思路分析

思路

• 用一个 dp 数组维护两个字符串的前缀编辑距离

• DP 定义

  • 记 word[0:i] := word 长度为 i 的 前缀子串
  • 定义 dp[i][j] := 将 word1[0:i] 转换为 word2[0:j] 的操作数

• 初始化

  dp[i][0] = i  // 每次从 word1 删除一个字符
  dp[0][j] = j  // 每次向 word1 插入一个字符
  

• 递推公式

  • word1[i] == word2[j]
    

    时

    dp[i][j] = dp[i-1][j-1]
    

  • word1[i] != word2[j]
    

    时，有三种更新方式，取最小

    // word[1:i] 表示 word 长度为 i 的前缀子串
    dp[i][j] = min({ dp[i-1][j]   + 1 ,     // 将 word1[1:i-1] 转换为 word2[1:j] 的操作数 + 删除 word1[i] 的操作数(1)
                     dp[i][j-1]   + 1 ,     // 将 word1[0:i] 转换为 word2[0:j-1] 的操作数 + 将 word2[j] 插入到 word1[0:i] 之后的操作数(1)
                     dp[i-1][j-1] + 1 })    // 将 word1[0:i-1] 转换为 word2[0:j-1] 的操作数 + 将 word1[i] 替换为 word2[j] 的操作数(1)
    

Code

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

Note

• 这道题难在思想上。