The regular expression matches the Roman number and returns the match to the value

The landlord of the two days to deal with a demand to match the Roman numerals in the string in a number of strings and take these Roman numbers for a series of operations. In the degree Niang on the search for a long day also did not find useful information also by a lot of wrong code misled the very painful. Fortunately, the last kinds of efforts It took half a day to reach the desired result. Now record and share it here. If you need to reprint please specify the source to write the portal. Thanks

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The following methods will take the Roman numerals in the string, but will also match the above string, so you need to post-process

/** * has been tested successfully take out regular expression matching string                 * @author Erikas        * @throws Exception */@org. junit.testpublic void Testreg () throws Ex ception {String str = "";//matches the regular of Roman numerals, but since each of them may be 0 empty strings will also be matched to be later in the program to process string regex = "(-| +|^) m{0,9} (cm| cd| D? c{0,3}) (xc| xl| L? x{0,3}) (ix|iv| V? i{0,3}) (+|$) "; Pattern p = pattern.compile (regex); Matcher Matcher = P.matcher (str); list<string> list = new arraylist<string> (); while (Matcher.find ()) {//matcher.find () returns true to match the result But after execution, if there is no subsequent match, the success will immediately become falseSystem.out.println ("here"); String srcstr = Matcher.group ();//Put the result of the removal into Listlist.add (SRCSTR);} SYSTEM.OUT.PRINTLN (list);}

The following methods are used to convert Roman numerals to corresponding Arabic numerals.

Roman numerals to Arabic numerals://Back to the    Roman numerals, if a number is smaller than the previous number, add that number to the result;    //vice versa, subtract the previous number two times in the result and add the current number;    //I, V, X,   L,   C,     D,     M    //1. 5, 10, 50, 100, 500, $    private static int r2a (String in) {        int graph[] = new int[400];        graph[' I '] = 1;        graph[' V ']=5;        graph[' X ']=10;        graph[' L ']=50;        graph[' C ']=100;        graph[' D ']=500;        graph[' M ']=1000;        char[] num = In.tochararray ();        Traverse this number, with Sum to total and        int sum = graph[num[0]];        for (int i=0; i<num.length-1; i++) {            //If, I is larger than i+1, add directly if            (Graph[num[i]] >= graph[num[i+1]]) {                Sum + = graph[num[i+1]];            }            If I is smaller than i+1, sum sum minus I this place number twice times, plus i+1//is equivalent to the number of the left side of the number is larger than the number of the left, then the numbers on the            right minus            else{                sum = sum + graph[num[i+ 1]]-2*graph[num[i]];            }        }        return sum;    }

The regular expression matches the Roman number and returns the match to the value