72. Edit Distance

Question

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')

Thinking:

• Method 1:dp dp[i][j]表示使从0开始到s1[i], s2[j]相同的最小步骤。
  • 初始化
    • 当i = 0时，dp[0][j] = j, 说明要添加j个元素才能使得s1[0], s2[j]相同。
    • 当j = 0时，dp[i][0] = i, 说明要从s1中删除i个元素才能变成s2[j].
  • 当求dp[i][j]时，取下列三个操作的最小值：
    • replace操作
      • s1[i] = s2[j]： dp[i - 1][j - 1], 说明前一位也是相同的，当前位也是相同的，不需要任何操作。
      • s1[i] != s2[j]: 当前位不相同，而dp[i - 1][j - 1]说明到前一位是相同的，我们只需要修改s1,s2的当前位。
    • dp[i - 1][j] + 1, s1的前一位已经和当前位相同了，我们需要删除当前的元素。
    • dp[i][j - 1] + 1, 说明s2的前一位已经和当前位相同了，我们需要在s1中插入s2的当前位。

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 + 1][0] = i + 1;
        for(int i = 0; i < len2; i++)
            dp[0][i + 1] = i + 1;
        for(int i = 1; i <= len1; i++){
            char c1 = word1.charAt(i - 1);
            for(int j = 1; j <= len2; j++){
                char c2 = word2.charAt(j - 1);
                if(c1 == c2){
                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 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];
    }
}

二刷

还是一刷时候的思路，如果面试应该能写出来。

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length(), len2 = word2.length();
        char[] arr1 = word1.toCharArray();
        char[] arr2 = word2.toCharArray();
        int[][] dp = new int[len1 + 1][len2 + 1];
        for(int i = 1; i <= len1; i++)
            dp[i][0] = i;
        for(int j = 1; j <= len2; j++)
            dp[0][j] = j;
        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                if(arr1[i - 1] == arr2[j - 1]){
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1] + 1), 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];
    }
}