LeetCode 72. Edit Distance

原题链接在这里：https://leetcode.com/problems/edit-distance/

题目：

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

题解：

 要求word1 convert成word2 需要的最少operations. 需要储存历史信息是word1到i段 convert成 word2 到j段需要的最少operations数目.

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

递推时,有两种情况，一种是word1 和 word2新检查的字符相同，那么operations数目dp[i][j]就没有变，还是dp[i-1][j-1].

若是字符不同的话就有三种操作:

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

2. 从word1减去1字符，operations数目是dp[i][j-1] +1;(可以理解成是向word2插入1字符，意思是相通的)

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

dp[i][j]取上面三个数字中最小的.

初始化一个从0 convert 成另一个到j的位置自然需要j次insert.

答案是dp[len1][len2].

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

AC Java:

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

只用到了dp[i-1][j-1]. dp[i-1][j] 和dp[i][j-1]三个数据. 可以降维节省空间.

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

AC Java:

class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null || word2 == null){
            throw new IllegalArgumentException("Invalid input strings.");
        }
        
        int len1 = word1.length();
        int len2 = word2.length();
        int [] dp = new int[len2+1];
        for(int j = 0; j<=len2; j++){
            dp[j] = j;
        }
        
        for(int i = 1; i<=len1; i++){
            int prev = i;
            for(int j = 1; j<=len2; j++){
                int cur;
                if(word1.charAt(i-1) == word2.charAt(j-1)){
                    cur = dp[j-1];
                }else{
                    int insert = dp[j]+1;
                    int delete = prev+1;
                    int replace = dp[j-1]+1;
                    cur = Math.min(insert, Math.min(delete, replace));
                }
                
                dp[j-1] = prev;
                prev = cur;
            }
            dp[len2] = prev;
        }
        return dp[len2];
    }
}

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