Leetcode calculation of string similarity or editing distance editdistance

Problem Description:

/**
 * Given, words word1 and word2, find the minimum number of steps required to
 * Convert Word1 to Word2. (Each operation is counted as 1 step.)
 * * You had the 
 following 3 operations permitted on a word:
 * 
 * a) Insert a character 
 * b) Delete a Chara Cter 
 * c) Replace a character
 */

The similarity of two strings is to calculate the number of steps required to convert a string into another string steps, inserting a character in one step, deleting one character as one step, and substituting one character for one step. So it's also called editing distance.
It is obvious that this is a dynamic programming problem. Declares an array arr[i][j] that represents the distance from the string word1[0...i-1] to word2[0...j-1].
There is a relationship between arr[i+1][j+1] and Arr[i][j], such as the word1 end character is the x,word2 end character is y:
If x=y, then arr[i+1][j+1]=arr[i][j];
If x!=y, then we can remove an element from the word1, arr[i+1][j+1]=arr[i+1][j]+1;
We can also insert an element from Word1, arr[i+1][j+1]=arr[i][j+1]+1;
We can also replace an element from Word1, arr[i+1][j+1]=arr[i][j]+1;
We can use the following figure to explain the above formula:

    A P p l e
  0 1 2 3 4 5
a 1 0 1 2 3 4
P 2 1 0 1 2 3
P 3 2 1 0 1 2

The first thing to initialize the arr array, arr[i][0]=i,arr[0][j]=j, is that the meaning of the initialization is arr[i][0]=i, that is, if the Word2 is empty, then word2 into the app, 0,1,2,3 steps are required, respectively. That is, word2 increases ' a ', increases ' p ', increases ' p '.
ARR[0][J] says that word1 from No to Word2 "Apple" needs 0,1,2,3,4,5 step, that is, increase ' a ', increase ' p ', increase ' p ', add ' l ', add ' e '.

One of Word1 's characters wants to be converted into WORD2, there are 3 ways: Replace, delete, add

Add 1 to each conversion, as this counts as 1 steps. This solves the above three formulas.

The specific code is as follows:

public int Mindistance (string word1, String word2) {if (word1.length () = = 0 | | word2.length () = = 0) r Eturn word1.length () = = 0?
        Word2.length (): Word1.length ();
        int[][] arr = new Int[word1.length () + 1][word2.length () + 1];
        for (int i = 0; I <= word1.length (); i++) {arr[i][0] = i;
        } for (int j = 0, J <= Word2.length (); j + +) {Arr[0][j] = j;
                } for (int i = 0; i < word1.length (), i++) {for (int j = 0; J < Word2.length (); + j) {
                if (Word1.charat (i) = = Word2.charat (j)) {arr[j + 1][i + 1] = Arr[j][i];
                    } else {int replace = arr[i][j]+1;
                    int insert = Arr[i][j + 1]+1;

                    int delete = Arr[i + 1][j]+1; int min = replace > insert?
                    Insert:replace; min = delete > min?
                    Min:delete; Arr[i + 1][j + 1] = min;
    }}} return Arr[word1.length ()][word2.length ()]; }