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Algorithm

72. Edit Distance

Description

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.

Solution

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        for(int i=0;i<=len1;i++){
            dp[i][0] = i;
        }
        for(int j=0;j<=len2;j++){
            dp[0][j] = j;
        }
        for(int i=1;i<=len1;i++){
            for(int j=1;j<=len2;j++){
                if(word1.charAt(i-1)==word2.charAt(j-1)){
                    // 两个字符完全相同
                    dp[i][j] = dp[i - 1][j - 1];
                }else{
                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]))+1;
                }
            }
        }
        return dp[len1][len2];
    }
}

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