72.Edit-Distance.md

72. Edit Distance

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题目地址

https://leetcode.com/problems/edit-distance/ https://www.jiuzhang.com/solutions/edit-distance

题目描述

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

state: f[i][j] word1的前i个字符最少要用几次编辑可以变成word2的前j个字符

function: f[i][j] 
= MIN(f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1]) // word1[i] == word2[j]
= MIN(f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1] + 1) // 
word1[i] != word2[j]

initialize: f[i][0] = i, f[0][j] = j

answer: f[a.length()][b.length()]

insert: f[i][j] = f[i][j-1] + 1
remove: f[i][j] = f[i-1][j] + 1
replace: f[i][j] = f[i-1][j-1] + 1

if (word1.charAt(i-1) == word2.charAt(j-1)){
  dp[i][j] = dp[i-1][j-1];
} else {
  dp[i][j] = dp[i-1][j-1] + 1;   //replace operation
}
dp[i][j] = Math.min(dp[i][j], Math.min(dp[i][j-1]+1, dp[i-1][j]+1));

代码

Approach #1 Dynamic Programming

从两个字符串的最后的位置开始考虑:

如果最后两个字符(i,j)相等，最后两个字符就不要配对，所以等于 minDistance(s1[0..i-1]，s2[0...j-1]) 如果最后两个字符不相等: 说明要编辑，具体可以分为三种情况:

• 如果 s1[i-1]和s2[j]可以配对，那我就删除s1[i]即可(删除)
• 如果 s1[i]和s2[j-1]可以配对，那我就在s1的后面加上s2[j]即可(插入)
• 如果 s1[i-1]和s2[j-1]可以配对，那我就把s1[i]修改成s2[j]即可

public class Solution {
    public int minDistance(String word1, String word2) {
        char[] s1 = word1.toCharArray();
        char[] s2 = word2.toCharArray();
        int i, j;
        int m = s1.length;
        int n = s2.length;

        int[][] f = new int[m + 1][n + 1];

        for (i = 0; i <= m; ++i) {
            for (j = 0; j <= n; ++j) {
                if (i == 0) {
                    f[i][j] = j;
                    continue;
                }

                if (j == 0) {
                    f[i][j] = i;
                    continue;
                }

                f[i][j] = Math.min(Math.min(f[i - 1][j], f[i][j - 1]), f[i - 1][j - 1]) + 1;

                if (s1[i - 1] == s2[j - 1]) {
                    f[i][j] = Math.min(f[i][j], f[i - 1][j - 1]);
                }

            }
        }

        return f[m][n];
    }
}

Approach #2

public class Solution {
  public int minDistance(String word1, String word2) {
    int n = word1.length();
    int m = word2.length();

    int[][] dp = new int[n + 1][m + 1];
    for (int i = 0; i < n + 1; i ++) {
        dp[i][0] = i;
    }
    for (int i = 0; i < m + 1; i ++) {
        dp[0][i] = i;
    }

    for (int i = 1; i < n + 1; i++) {
      for (int j = 1; j < m + 1; j++) {
        if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
            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[n][m];
  }
}