LeetCode 72. Edit Distance 编辑距离 (C++/Java)

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')

分析：

给定两个单词，求word1转换到word2须要的最少步骤，转换的操做有三种，分别是插入一个字符，删除一个字符，替换一个字符。

d(word1, word2)用来求两个单词转换的须要的最小步数，那么若是当两个单词的最后一个字符是相同的，则d(word1, word2) = d(word1', word2')其word1'和word2'是分别去掉最后一个字符的单词。

若是最后两个字符不相同时，咱们就须要操做来进行转换，一种是在word1后增长一个字符，是其最后一个字符和word2的最后一个字符相同，一种是删去word1的最后一个字符，一种是将word1的最后一个字符转换成word2的最后一个字符，那么此时最小的步数就是前三个操做的最小值加上1.

可能有同窗会问为何不在word2上进行操做，实际上操做转换这一步是有5个子问题的，但实际上在word1后增长一个字符和word2最后字符相同，至关于在word2后删除字符；删去word1的字符至关于在word2后增长一个字符和word1最后字符相同；而转换操做明显是同样的，因此就合并成为了三个子问题。

当递归执行到其中一个串为空串时，则加上另外一个串的长度便可，至关于删去全部的字符。

程序：

C++

class Solution { public: int minDistance(string word1, string word2) { int l1 = word1.length(); int l2 = word2.length(); dp = vector<vector<int>>(l1+1, vector<int>(l2+1, -1)); return minDistance(word1, word2, l1, l2); } private: vector<vector<int>> dp; int minDistance(string& word1, string& word2, int l1, int l2){ if(l1 == 0) return l2; if(l2 == 0) return l1; if(dp[l1][l2] >= 0) return dp[l1][l2]; int res = 0; if(word1[l1-1] == word2[l2-1]){ res = minDistance(word1, word2, l1-1, l2-1); dp[l1][l2] = res; return res; } res = min(minDistance(word1, word2, l1-1, l2), min(minDistance(word1, word2, l1, l2-1), minDistance(word1, word2, l1-1, l2-1))) + 1; dp[l1][l2] = res; return res; } };

Java

class Solution { public int minDistance(String word1, String word2) { int l1 = word1.length(); int l2 = word2.length(); dp = new int[l1+1][l2+1]; for(int i = 0; i < dp.length; ++i){ for(int j = 0; j < dp[i].length; ++j){ dp[i][j] = -1; } } return minDistance(word1, word2, l1, l2); } private int minDistance(String word1, String word2, int l1, int l2){ if(l1 == 0) return l2; if(l2 == 0) return l1; if(dp[l1][l2] >= 0) return dp[l1][l2]; int res = 0; if(word1.charAt(l1-1) == word2.charAt(l2-1)){ res = minDistance(word1, word2, l1-1, l2-1); }else{ res = Math.min(minDistance(word1, word2, l1-1, l2), Math.min(minDistance(word1, word2, l1, l2-1), minDistance(word1, word2, l1-1, l2-1))) + 1; } dp[l1][l2] = res; return res; } private int[][] dp; }