为什么不采取天真的方式呢？

bool matchesAlmost(String str1, String str2) {
    if (str1.length != str2.length)
        return false;
    int same = 0;
    for (int i = 0; i < str1.length; ++i) {
        if (str1.charAt(i) == str2.charAt(i))
            same++;
    }
    return same == str1.length - 1;
}

现在你可以用一个二次算法来检查每一个字符串。

假设两个字符串的长度相等

String str1 = "rock";
        String str2 = "lick";

        if( str1.length() != str2.length() )
            System.out.println( "failed");

        else{
            if( str2.contains( str1.substring( 0, str1.length()-1)) || str2.contains(  str1.substring(1, str1.length() )) ){
                System.out.println( "Success ");
            }

            else{
                System.out.println( "Failed");
            }
        }

不确定这是否是最好的方法，但即使两个字符串长度不同，这一方法也能工作。cat & cattp它们只差一个字符p，并且t是重复的。看起来像是O（n）时间的解决方案，使用额外的空间来存放散列表和字符数组。

/**
 * Returns true if two strings differ by one character
 * @param s1 input string1
 * @param s2 input string2
 * @return true if strings differ by one character 
 */
boolean checkIfTwoStringDifferByOne(String s1, String s2) {
    char[] c1, c2;

    if(s1.length() < s2.length()){
        c1 = s1.toCharArray();
        c2 = s2.toCharArray();
    }else{
        c1 = s2.toCharArray();
        c2 = s1.toCharArray();
    }

    HashSet<Character> hs = new HashSet<Character>();

    for (int i = 0; i < c1.length; i++) {
        hs.add(c1[i]);
    }
    int count = 0;
    for (int j = 0; j < c2.length; j++) {
        if (! hs.contains(c2[j])) {
            count = count +1;
        }
    }

    if(count == 1)
        return true;
    return false;
}

假设所有字符串的长度都相同，我想这会有所帮助：

public boolean differByOne(String source, String destination)
{
    int difference = 0;

    for(int i=0;i<source.length();i++)
    {
        if(source.charAt(i)!=destination.charAt(i))
        {
            difference++;

            if(difference>1)
            {
                return false;
            }
        }
    }

    return difference == 1;
}

最好的方法是将字符串连接在一起，一个向前，另一个以相反的顺序。然后检查两端匹配的字符，并从中间开始向两端匹配的字符单循环。如果超过2个字符不匹配中断。
如果一个不匹配，则停止并等待下一个完成，如果它到达相同的位置，则它匹配，否则只返回false。

public static void main(String[] args) {
    New1 x = new New1();
    x.setFunc();
}

static void setFunc() {
    Set s = new HashSet < Character > ();
    String input = " aecd";
    String input2 = "abcd";
    String input3 = new StringBuilder(input2).reverse().toString();
    String input4 = input.concat(input3);
    int length = input4.length();

    System.out.println(input4);
    int flag = 0;

    for (int i = 1, j = length - 1; j > i - 1; i++, j--) {

        if (input4.charAt(i) != input4.charAt(j)) {

            System.out.println(input4.charAt(i) + " doesnt match with " + input4.charAt(j));
            if (input4.charAt(i + 1) != input4.charAt(j)) {
                System.out.println(input4.charAt(i + 1) + " doesnt match  with " + input4.charAt(j));
                flag = 1;
                continue;
            } else if (input4.charAt(i) != input4.charAt(j - 1)) {
                System.out.println(input4.charAt(i) + " doesnt match with " + input4.charAt(j - 1));
                flag = 1;
                break;
            } else if (input4.charAt(i + 1) != input4.charAt(j - 1) && i + 1 <= j - 1) {
                System.out.println(input4.charAt(i + 1) + " doesnt match with xxx " + input4.charAt(j - 1));
                flag = 1;
                break;
            }
        } else {
            continue;
        }
    }

    if (flag == 0) {
        System.out.println("Strings differ by one place");
    } else {
        System.out.println("Strings does not match");
    }
}