Question :
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
Anwser 1 :
class Solution {
public:
vector<string> generateParenthesis(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<string> vec;
string str;
addParen(vec,str, n, n);
return vec;
}
void addParen(vector<string> &vec, string str, int left, int right) {
if (left == 0 && right == 0) {
vec.push_back(str);
return;
}
if (left > 0) {
addParen(vec, str + '(', left - 1, right);
}
if (right > left) {
addParen(vec, str + ')', left, right-1);
}
}
};
Anwser 2 :
class Solution {
public:
void printPar(int l, int r, vector<string>& result, char* str, int idx)
{
if (l < 0 || r < l) return;
if (l == 0 && r == 0)
{
str[idx] = '\0';
result.push_back(string(str));
}
else
{
if (l > 0)
{
str[idx] = '(';
printPar(l-1, r, result, str, idx+1);
}
if (r > l)
{
str[idx] = ')';
printPar(l, r-1, result, str, idx+1);
}
}
}
vector<string> generateParenthesis(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<string> result;
char* str = new char[2*n+1];
printPar(n, n, result, str, 0);
delete []str;
return result;
}
};
Anwser 3 :
class Solution {
public:
vector<string> generateParenthesis(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<string> ans;
if (n > 0) {
generator(ans, "", 0, 0, n);
}
return ans;
}
// r/l: appearance of ) (
void generator(vector<string> & ans, string s, int l, int r, int n) {
if (l == n) {
ans.push_back(s.append(n-r, ')'));
return;
}
generator(ans, s + '(', l+1, r, n);
if (l > r) {
generator(ans, s + ")", l, r+1, n);
}
}
};
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