0072. Edit Distance (H)

Edit Distance (H)

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:

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')

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题意

提供三种操作：插入一个字符，删除一个字符，替换一个字符，要求使用最少的操作将一个字符串转变为另一个字符串。

思路

动态规划。(dp[i][j])表示将(S[0, i - 1])转变为(T[0, j - 1])所需的最少操作数。每次比较S和T的最后一个字符，两种情况：

1. (S[i - 1] == T[j - 1])。说明只要将(S[0, i - 2])转变为(T[0, j - 2])，有(dp[i][j]=dp[i-1][j-1])

2. (S[i-1] != T[j-1])。为了使最后一位相同，有三种操作：

   1. 将S的最后一位替换为T的最后一位，有(dp[i][j]=dp[i-1][j-1]+1)
   2. 将S的最后一位删去，有(dp[i][j]=dp[i-1][j]+1)
   3. 将T的最后一位插入到S的最后，有(dp[i][j]=dp[i][j-1]+1)

   取三种情况中的最小值作为(dp[i][j])。

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代码实现

Java

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length(), 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)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
                }
            }
        }
        return dp[len1][len2];
    }
}