问题
Hi I have two strings :
String hear = \"Hi My name is Deepak\"
+ \"\\n\"
+ \"How are you ?\"
+ \"\\n\"
+ \"\\n\"
+ \"How is everyone\";
String dear = \"Hi My name is Deepak\"
+ \"\\n\"
+ \"How are you ?\"
+ \"\\n\"
+ \"Hey there \\n\"
+ \"How is everyone\";
I want to get what is not present in the hear string that is \"Hey There \\n\". I found a method , but it fails for this case :
static String strDiffChop(String s1, String s2) {
if (s1.length() > s2.length()) {
return s1.substring(s2.length() - 1);
} else if (s2.length() > s1.length()) {
return s2.substring(s1.length() - 1);
} else {
return \"\";
}
}
Can any one help ?
回答1:
google-diff-match-patch
The Diff Match and Patch libraries offer robust algorithms to perform the operations required for synchronizing plain text.
Diff:
Compare two blocks of plain text and efficiently return a list of differences.
Match:
Given a search string, find its best fuzzy match in a block of plain text. Weighted for both accuracy and location.
Patch:
Apply a list of patches onto plain text. Use best-effort to apply patch even when the underlying text doesn't match.
Currently available in Java, JavaScript, Dart, C++, C#, Objective C, Lua and Python. Regardless of language, each library features the same API and the same functionality. All versions also have comprehensive test harnesses.
There is a Line or word diffs wiki page which describes how to do line-by-line diffs.
回答2:
One can use the StringUtils from Apache Commons. Here is the StringUtils API.
public static String difference(String str1, String str2) {
if (str1 == null) {
return str2;
}
if (str2 == null) {
return str1;
}
int at = indexOfDifference(str1, str2);
if (at == -1) {
return EMPTY;
}
return str2.substring(at);
}
public static int indexOfDifference(String str1, String str2) {
if (str1 == str2) {
return -1;
}
if (str1 == null || str2 == null) {
return 0;
}
int i;
for (i = 0; i < str1.length() && i < str2.length(); ++i) {
if (str1.charAt(i) != str2.charAt(i)) {
break;
}
}
if (i < str2.length() || i < str1.length()) {
return i;
}
return -1;
}
回答3:
I have used the StringTokenizer to find the solution. Below is the code snippet
public static List<String> findNotMatching(String sourceStr, String anotherStr){
StringTokenizer at = new StringTokenizer(sourceStr, " ");
StringTokenizer bt = null;
int i = 0, token_count = 0;
String token = null;
boolean flag = false;
List<String> missingWords = new ArrayList<String>();
while (at.hasMoreTokens()) {
token = at.nextToken();
bt = new StringTokenizer(anotherStr, " ");
token_count = bt.countTokens();
while (i < token_count) {
String s = bt.nextToken();
if (token.equals(s)) {
flag = true;
break;
} else {
flag = false;
}
i++;
}
i = 0;
if (flag == false)
missingWords.add(token);
}
return missingWords;
}
回答4:
convert the string to lists and then use the following method to get result How to remove common values from two array list
回答5:
If you prefer not to use an external library, you can use the following Java snippet to efficiently compute the difference:
/**
* Returns an array of size 2. The entries contain a minimal set of characters
* that have to be removed from the corresponding input strings in order to
* make the strings equal.
*/
public String[] difference(String a, String b) {
return diffHelper(a, b, new HashMap<>());
}
private String[] diffHelper(String a, String b, Map<Long, String[]> lookup) {
return lookup.computeIfAbsent(((long) a.length()) << 32 | b.length(), k -> {
if (a.isEmpty() || b.isEmpty()) {
return new String[]{a, b};
} else if (a.charAt(0) == b.charAt(0)) {
return diffHelper(a.substring(1), b.substring(1), lookup);
} else {
String[] aa = diffHelper(a.substring(1), b, lookup);
String[] bb = diffHelper(a, b.substring(1), lookup);
if (aa[0].length() + aa[1].length() < bb[0].length() + bb[1].length()) {
return new String[]{a.charAt(0) + aa[0], aa[1]};
} else {
return new String[]{bb[0], b.charAt(0) + bb[1]};
}
}
});
}
This approach is using dynamic programming. It tries all combinations in a brute force way but remembers already computed substrings and therefore runs in O(n^2).
Examples:
String hear = "Hi My name is Deepak"
+ "\n"
+ "How are you ?"
+ "\n"
+ "\n"
+ "How is everyone";
String dear = "Hi My name is Deepak"
+ "\n"
+ "How are you ?"
+ "\n"
+ "Hey there \n"
+ "How is everyone";
difference(hear, dear); // returns {"","Hey there "}
difference("Honda", "Hyundai"); // returns {"o","yui"}
difference("Toyota", "Coyote"); // returns {"Ta","Ce"}
回答6:
I was looking for some solution but couldn't find the one i needed, so I created a utility class for comparing two version of text - new and old - and getting result text with changes between tags - [added] and [deleted]. It could be easily replaced with highlighter you choose instead of this tags, for example: a html tag. string-version-comparison
Any comments will be appreciated.
*it might not worked well with long text because of higher probability of finding same phrases as deleted.
回答7:
what about this snippet ?
public static void strDiff(String hear, String dear){
String[] hr = dear.split("\n");
for (String h : hr) {
if (!hear.contains(h)) {
System.err.println(h);
}
}
}
回答8:
You should use StringUtils from Apache Commons
String diff = StringUtils.difference( "Word", "World" );
System.out.println( "Difference: " + diff );
Difference: ld
Source: https://www.oreilly.com/library/view/jakarta-commons-cookbook/059600706X/ch02s15.html
来源：https://stackoverflow.com/questions/18344721/extract-the-difference-between-two-strings-in-java