尝试将C清晰代码转换为Java时出现问题

Question

i'm try find most similar string in a array, and i found a code in c sharp that is this one

  public static int LevenshteinDistance(string s, string t)
    {
        int n = s.Length;
        int m = t.Length;
        int[,] d = new int[n + 1, m + 1];
        if (n == 0)
        {
            return m;
        }
        if (m == 0)
        {
            return n;
        }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
                Console.WriteLine(cost);
                d[i, j] = Math.Min(
                    Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
                    d[i - 1, j - 1] + cost);


            }
        }
        return d[n, m];
    }

and i'm trying to convert it into java but i get 1 error this is my code in java

    public static int LevenshteinDistance(String s, String t)
       {
         int n = s.length();
        int m = t.length();
        int[][] d = new int[n + 1][ m + 1];
        if (n == 0)
        {
            return m;
        }
        if (m == 0)
        {
            return n;
        }



        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {


            int cost = (t[j - 1] == s[i - 1])? 0 : 1;

                d[i][ j] = Math.min(
                    Math.min(d[i - 1][ j] + 1, d[i][ j - 1] + 1),
                    d[i - 1][ j - 1]+cost );
            }
        }
        return d[n] [m];
    }

i get the error in this line of code int cost = (t[j - 1] == s[i - 1]) ? 0 : 1; the error that i have is "Array is required,but string found" this is what i have in my main

         String []ray ={"food","drinks","stuffs"};
         String fa="drink";
        for (int i = 0; i < ray.length; i++)
        {

         System.out.print(LevenshteinDistance(fa, ray[i]));
        }

i would appreciate any help

Answer 1

使用t.charAt（j-1）== s.charAt（i-1）来访问字符串中的字符（字母）您不能直接通过索引（括号[]）访问它们。

int cost = (t.charAt(j - 1) == s.charAt(i - 1))? 0 : 1;

Answer 2

You are accessing the strings as arrays here with the [] array operator:

t[j - 1] == s[i - 1]

to get the nth char of a string, instead use .charAt(n) so in this case change it to:

t.charAt(j - 1) == s.charAt(i - 1)

the same applies to the rest of the code.