"Leetcode" Generate parentheses (2 solutions)

Generate parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

Solution One: Recursion

With Stacks, ' (', ') ' constitutes a pair of separate entry and exit stacks. When the last stack is empty, the input parentheses make up a string that is legal.

classSolution { Public: Vector<string> Generateparenthesis (intN) {vector<string>result; if(n = =0)            returnresult; //First must be ' ('        stringCur ="("; Stack<Char>s; S.push ('('); Helper (result, cur, s,2*n-1); returnresult; }    voidHelper (vector<string>& result,stringCur, stack<Char> S,intnum) {        if(num = =1)        {//must be ') '            if(S.top () = ='('&& s.size () = =1)            {//All matchedCur + =')';            Result.push_back (cur); }        }        Else        {            //' (' Always push            stringSTR1 =cur; STR1+='('; S.push ('('); Helper (result, STR1, S, Num-1);                        S.pop (); //' ) '            if(!S.empty ()) {//prune. Never begin with ') '                stringSTR2 =cur; STR2+=')'; if(S.top () = ='(') S.pop (); //check Empty () before access top ()                ElseS.push (')'); Helper (result, STR2, S, Num-1); }        }    }};

Solution Two: Recursion

A little analysis shows that the stack is unnecessary, as long as there are several ' (', Count ') records in the string.

Each entry into a ' (', Count + +.) Each matching pair of parentheses, Count--。

Eventually all match, need count==0

classSolution { Public: Vector<string> Generateparenthesis (intN) {vector<string>result; if(n = =0)            returnresult; //First must be ' ('        stringCur ="("; intCount =1;//Number of (' s in curHelper (result, cur, count,2*n-1); returnresult; }    voidHelper (vector<string>& result,stringCurintCountintnum) {        if(num = =1)        {//must be ') '            if(Count = =1)            {//All matchedCur + =')';            Result.push_back (cur); }        }        Else        {            //' (' Always push            stringSTR1 =cur; STR1+='('; Count++; Helper (result, str1, count, Num-1); Count--; //' ) '            if(Count! =0)            {//prune. Never begin with ') '                stringSTR2 =cur; STR2+=')'; Count--; Helper (result, str2, count, Num-1); }        }    }};

"Leetcode" Generate parentheses (2 solutions)