Matching brackets (string)

Problem description:
Check whether the parentheses in the string expression match;
The number of left parentheses is not matched if the number of parentheses is not equal;
Remove the Left or Right brackets and keep the parentheses first;
Remove invalid parentheses after matching: (expression) should be (expression );
Only parentheses are considered, not the case where parentheses appear first;
Required implementation function: (the maximum length of a string is 60. You do not need to consider whether the expression is correct)
Void bracket (char * SRC, char * DST );
If yes, the original string is output through DST;
If the parameter does not match, the number of matched strings is greater than that of the string after the brackets are output through DST.
Example
Input: 12 + (345*25-34) Output: 12 + (345*25-34)
Input: 12 + (345 * (25-34) Output: 12 + (345*25-34)
Input: (12 + 345) * 25)-34 output: (12 + 345) * 25-34
Input: (543 + (256-43) * 203) + 24
Output: (543 + (256-43) * 203) + 24
Input: (1 + 2) * (34-2) + (2*8-1)
Output: (1 + 2) * (34-2) + 2*8-1)

# Include <iostream> # include <vector> # include <string> # include <algorithm> using namespace STD; void leftright (string S, vector <int> & leftbracket, vector <int> & rightbracket) {unsigned Len = S. size (); leftbracket. clear (); rightbracket. clear (); For (unsigned I = 0; I <Len; I ++) {If (s [I] = '(') leftbracket. push_back (I); else if (s [I] = ') rightbracket. push_back (I) ;}} void bracket (char * SRC, char * DST) {vector <int> L Eftbracket, rightbracket; // subscript unsigned Len = strlen (SRC); string S (SRC, SRC + Len); leftright (S, leftbracket, rightbracket ); unsigned len1 = leftbracket. size (), len2 = rightbracket. size (); While (len1> len2) {S. erase (leftbracket [len1-1], 1); leftright (S, leftbracket, rightbracket); len1 --;} while (len1 <len2) {S. erase (rightbracket [len2-1], 1); leftright (S, leftbracket, rightbracket); len2 --;} unsigned K1, K2; vector <int> era; // subscript for the element to be deleted (k1 = len1-1; k1> 0; K1 --) // Delete the undesired bracket pair {If (leftbracket [k1]-leftbracket [k1-1]! = 1) {for (k2 = 0; K2 <len2; K2 ++) {If (leftbracket [k1] <rightbracket [k2]) {rightbracket [k2] = 0; break;} // end for} // end if else {for (k2 = 0; K2 <len2-1; K2 ++) {If (leftbracket [k1] <rightbracket [k2] & leftbracket [k1-1] <rightbracket [k2 + 1] & rightbracket [k2 + 1]-rightbracket [k2] = 1) {era. push_back (leftbracket [k1]); era. push_back (rightbracket [k2]); rightbracket [k2] = 0; break ;}// end for}/end else}/E Nd forif (era. Size ()! = 0) // Delete redundant arc {sort (era. begin (), era. end (); For (INT I = era. size ()-1; I> = 0; I --) {S. erase (ERA [I], 1) ;}} strncpy (DST, & S [0], S. size (); DST [S. size ()] = '\ 0';} void main () {char * src = "(1 + 2) * (34-2 )) + (2*8-1) "; char * DST = new char [100]; bracket (SRC, DST); puts (DST );}

Note:

If the erase function of string is used, the correspondence between the size of string and the previous subscript and the string is changed. Pay special attention to this programming!