java - 如何确定一个字符串是否只在java中编辑过一次？

Question

我正在尝试确定一个字符串是否只编辑过一次。当我输入以下字符串（见下面的代码）时，输出不正确。程序应该打印出字符串只编辑过一次，这不是我在屏幕上看到的。谁能帮帮我？

public static void main(String[] args) {
        String str1 = "cat";
        String str2 = "ca";
        char[] arr1 = str1.toCharArray();
        char[] arr2 = str2.toCharArray();
        int n = str1.length();
        int m = str2.length();
        if (stringWasEdited(str1, str2, arr1, arr2, m, n)) {
            System.out.print("The string was edited only once.");
        } else {
            System.out.print("The string was not edited or was edited more than once.");
        }
    }

    private static boolean stringWasEdited(String str1, String str2, char[] arr1, char[] arr2, int m, int n) {
        int count = 0;
        if (Math.abs(m - n) > 1) {
            return false;
        }
        for (int i = 0; i < m; i++) {
            if (arr1[i] != arr2[i]) {
                count++;
            }
            if (m > n || n > m) {
                count++;
            }
        }
        if (count > 1 || count == 0) {
            return false;
        }
        return true;
    }
}

Answer 1

你的问题是你只循环一个字符串的长度，而你应该从 i

static boolean isEditDistanceOne(String s1,
                                 String s2)
{
    // Find lengths of given strings
    int m = s1.length(), n = s2.length();
 
    // If difference between lengths is
    // more than 1, then strings can't
    // be at one distance
    if (Math.abs(m - n) > 1)
        return false;
 
    int count = 0; // Count of edits
 
    int i = 0, j = 0;
    while (i < m && j < n)
    {
        // If current characters don't match
        if (s1.charAt(i) != s2.charAt(j))
        {
            if (count == 1)
                return false;
 
            // If length of one string is
            // more, then only possible edit
            // is to remove a character
            if (m > n)
                i++;
            else if (m< n)
                j++;
            else // Iflengths of both strings
                // is same
            {
                i++;
                j++;
            }
             
            // Increment count of edits
            count++;
        }
 
        else // If current characters match
        {
            i++;
            j++;
        }
    }
 
    // If last character is extra
    // in any string
    if (i < m || j < n)
        count++;
 
    return count == 1;
}

参考链接：https://www.geeksforgeeks.org/check-if-two-given-strings-are-at-edit-distance-one/

Answer 2

我假设您想检查一个编辑，拒绝包含零个或多个的字符串。

给定两个字符串 (s1, s2)，其中 s1 的长度大于或等于 s2（如果不是这种情况，请反转参数）我们可以使用以下命令检查单个编辑：

1. 过滤掉长度差大于1的情况。

2. 使用第一个相等字符的序列。

3. 如果您已到达s2 的末尾，请检查s1 是否更长。如果不是，则s1 == s2 不满足条件。

4. 使用第二个相等字符序列。如果s1 更长（加一），那么我们在访问s2 中的字符时需要使用偏移量1。

5. 如果您已到达s1 的末尾，则满足条件。

以上翻译成Java（Ideone）：

static boolean hasOneEdit(String s1, String s2)
{
    // ensure s1 is longer than or equal to s2
    if(s1.length() < s2.length()) return hasOneEdit(s2, s1);
    
    // difference in length cannot be more than one
    if(s1.length() - s2.length() > 1) return false;

    int pos = 0;
    
    // consume 1st sequence of equal characters
    for(; pos < s2.length() && s1.charAt(pos) == s2.charAt(pos); pos++); 
    
    // if we reached the end of s2, s1 has to be longer (by 1)
    if(pos == s2.length()) return pos < s1.length();

    // offset into s2 depends on whether strings are equal in length
    int off = s1.length() - s2.length();
            
    // consume 2nd sequence of equal characters
    for(pos += 1; pos < s1.length() && s1.charAt(pos) == s2.charAt(pos-off); pos++);
    
    // did we reach the end of s1?
    return pos == s1.length();
}

还有一个小测试程序：

public static void main(String[] args)
{
    String s1 = "dirigible";
    for(int i=0; i<s1.length(); i++)
    {
        String s2 = s1.substring(0, i) + s1.substring(i+1);
        test(s1, s2, true);
        for(int j=0; j<s2.length(); j++)
        {
            String s3 = s2.substring(0, j) + "x" + s2.substring(j+1);
            test(s1, s3, false);
        }               
    }
    for(int i=0; i<s1.length(); i++)
    {
        String s2 = s1.substring(0, i) + "x" + s1.substring(i+1);
        test(s1, s2, true);
        for(int j=0; j<s2.length(); j++)
        {               
            String s3 = s2.substring(0, j) + s2.substring(j+1);
            test(s1, s3, j == i);
        }
    }
}

static void test(String s1, String s2, boolean check)
{
    boolean result = hasOneEdit(s1, s2);
    System.out.println((result == check ? "PASS" : "FAIL") + ": " + s1 + " " + s2 + " = " + check);
}

输出：

PASS: dirigible irigible = true
PASS: dirigible xrigible = false
PASS: dirigible ixigible = false
PASS: dirigible irxgible = false
PASS: dirigible irixible = false
<Truncated>
PASS: dirigible dirgiblx = false
PASS: dirigible diriiblx = false
PASS: dirigible dirigblx = false
PASS: dirigible dirigilx = false
PASS: dirigible dirigibx = false
PASS: dirigible dirigibl = true