LeetCode 72. Edit Distance

原題連接在這裏：https://leetcode.com/problems/edit-distance/

題目：

Given 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

題解：

 要求word1 convert成word2 須要的最少operations. 須要儲存歷史信息是word1到i段 convert成 word2 到j段須要的最少operations數目.

Let dp[i][j] denotes the minimum number of operations to convert from word1 till index i to word2 till index j. 

遞推時,有兩種狀況，一種是word1 和 word2新檢查的字符相同，那麼operations數目dp[i][j]就沒有變，仍是dp[i-1][j-1].

如果字符不一樣的話就有三種操做:

1. 向word1插入1字符，operations數目是dp[i-1][j] + 1;

2. 從word1減去1字符，operations數目是dp[i][j-1] +1;(能夠理解成是向word2插入1字符，意思是相通的)

3. word1 replace 1字符, operations數目是dp[i-1][j-1]+1.

dp[i][j]取上面三個數字中最小的.

初始化一個從0 convert 成另外一個到j的位置天然須要j次insert.

答案是dp[len1][len2].

Time Complexity: O(len1 * len2). Space: O(len1 * len2).

AC Java:

 1 public class Solution {
 2     public int minDistance(String word1, String word2) {
 3         if(word1 == null || word2 == null){
 4             return -1;
 5         }
 6         int len1 = word1.length();
 7         int len2 = word2.length();
 8         int [][] dp = new int[len1+1][len2+1];
 9         for(int i=0; i<=len1; i++){
10             dp[i][0] = i;
11         }
12         for(int j=0; j<=len2; j++){
13             dp[0][j] = j;
14         }
15         for(int i=1; i<=len1; i++){
16             for(int j=1; j<=len2; j++){
17                 if(word1.charAt(i-1) == word2.charAt(j-1)){
18                     dp[i][j] = dp[i-1][j-1];
19                 }else{
20                     int insert = dp[i-1][j] + 1;
21                     int delete = dp[i][j-1] + 1;
22                     int replace = dp[i-1][j-1] + 1;
23                     dp[i][j] = Math.min(insert, Math.min(delete,replace));
24                 }
25             }
26         }
27         return dp[len1][len2];
28     }
29 }

只用到了dp[i-1][j-1]. dp[i-1][j] 和dp[i][j-1]三個數據. 能夠降維節省空間.

Time Complexity: O(len1*len2). Space: O(len2).

AC Java:

 1 class Solution {
 2     public int minDistance(String word1, String word2) {
 3         if(word1 == null || word2 == null){
 4             throw new IllegalArgumentException("Invalid input strings.");
 5         }
 6         
 7         int len1 = word1.length();
 8         int len2 = word2.length();
 9         int [] dp = new int[len2+1];
10         for(int j = 0; j<=len2; j++){
11             dp[j] = j;
12         }
13         
14         for(int i = 1; i<=len1; i++){
15             int prev = i;
16             for(int j = 1; j<=len2; j++){
17                 int cur;
18                 if(word1.charAt(i-1) == word2.charAt(j-1)){
19                     cur = dp[j-1];
20                 }else{
21                     int insert = dp[j]+1;
22                     int delete = prev+1;
23                     int replace = dp[j-1]+1;
24                     cur = Math.min(insert, Math.min(delete, replace));
25                 }
26                 
27                 dp[j-1] = prev;
28                 prev = cur;
29             }
30             dp[len2] = prev;
31         }
32         return dp[len2];
33     }
34 }

相似One Edit Distance, Delete Operation for Two Strings, Minimum ASCII Delete Sum for Two Strings, Longest Common Subsequence, Regular Expression Matching.