题目
Edit Distance
Total Accepted: 9568 Total Submissions: 38449My SubmissionsGiven two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
解法
借用这道题来讨论一下编辑距离这个话题。思路
微软实习在2014年春天第一次网上提交代码时候，曾经出过一道跟编辑距离有关的题目。 拿到这道题之前，我们的思路是：- 了解什么是编辑距离
- 编辑距离建模
- 因为是DP系列题目，所以跳过一些判断直接想能否用DP解
编辑距离
我们首先想到的是算法导论上关于编辑距离的定义，但是在此之前我们先看看wikipedia上面的描述：In computer science, edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another by counting the minimum number of operations required to transform one string into the other. Edit distances find applications in natural language processing, where automatic spelling correction can determine candidate corrections for a misspelled word by selecting words from a dictionary that have a low distance to the word in question. Inbioinformatics, it can be used to quantify the similarity of macromolecules such as DNA, which can be viewed as strings of the letters A, C, G and T.
Given two strings a and b on an alphabet Σ (e.g. the set of ASCII characters, the set of bytes [0..255], etc.), the edit distance d(a, b) is the minimum-weight series of edit operations that transforms a into b. One of the simplest sets of edit operations is that defined by Levenshtein in 1966:
- Insertion of a single symbol. If a = uv, then inserting the symbol x produces uxv. This can also be denoted ε→x, using ε to denote the empty string.
- Deletion of a single symbol changes uxv to uv (x→ε).
- Substitution of a single symbol x for a symbol y ≠ x changes uxv to uyv (x→y).
由于这个是一道跟字符串有关的题目，按照LeetCode传统的思路就是按字符比较。我们再看看wikipedia上面的一个例子（是关于了是否有substitution的例子）
The Levenshtein distance between "kitten" and "sitting" is 3. The minimal edit script that transforms the former into the latter is:
- kitten → sitten (substitution of "s" for "k")
- sitten → sittin (substitution of "i" for "e")
- sittin → sitting (insertion of "g" at the end).
LCS distance(insertions and deletions only) gives a different distance and minimal edit script:
- delete k at 0
- insert s at 0
- delete e at 4
- insert i at 4
- insert g at 6
for a total cost/distance of 5 operations.
然后我们考虑一下传统DP的解法（算导中的一节，为什么做DP就不详细说）。
DP解法
DP操作建模
对于编辑距离的三个操作来说：插入举例：X：=ex VS Y:=exp前面的两个ex都是相同，则编辑距离不变为0，X没有第三个字符，这里如果我们插入了一个字符即可。删除举例： X：= exp VS Y:= ex我故意用了这个例子，也就是说插入和删除是互逆的，就是看我们以谁为对象来看待这个问题。
替代举例： X:= sitten VS Y:=sitting这里例子的第五个字符需要替代，为什么要替代呢？ 因为第四个字符和第六个字符相同（这里就需要用DP，也就是说，需要一个空间来记录之前的比较结果和之后的比较结果）。
DP实现：
考虑了几个例子之后，我们再想一下这三个步骤是同时判断的，还是有先后顺序来判断的？我们再看看wikipedia里面的基本的算法：Basic algorithmMain article: Wagner–Fischer algorithm
Using Levenshtein's original operations, the edit distance between
and
is given by 
, defined by the recurrence
This algorithm can be generalized to handle transpositions by adding an additional term in the recursive clause's minimization
public 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];
if(len1 == 0) return len2;
if(len2 == 0) return len1;
for(int i = 0; i < len1+1; i++)
dp[i][0] = i;
for(int i = 0 ; i < len2+1; i++)
dp[0][i] = i;
for(int i = 1; i < len1+1; i++){
for(int j = 1 ; j <len2+1;j++){
//dp[i][j] VS dp[i][j-1]+1(cost) for delete VS dp[i-1][j] +1(cost) for insert VS dp[i-1][j-1] + cost for substitution
int cost = word1.charAt(i-1)==word2.charAt(j-1) ? 0 : 1;
dp[i][j] = Math.min(dp[i-1][j-1]+cost,Math.min(dp[i][j-1]+1,dp[i-1][j]+1));
}
}
return dp[len1][len2];
}
}
博客参考：wikipedia博客查阅：
[leetcode]Edit Distance
代码优化：[Leetcode] Edit Distance进阶学习：自然语言处理学习篇02——Edit Distance